Here’s a theorem from O’Neill’s Semi-Riemannian Geometry, With Applications to Relativity:
Theorem (O’Neill, Chapter 5, Lemma 37).
For a smooth manifold $M$, the following statements are equivalent:
- $M$ admits a Lorentzian metric
- $M$ admits a time-orientable Lorentzian metric
- $M$ admits a smooth nowhere-vanishing vector field
- $M$ is either non-compact or it is compact with vanishing Euler-characteristic: $\chi(M)=0$.
The proof is relatively short once you accept the equivalence of (3) and (4) by referring to some other source. So, the question of smooth manifolds admitting (time-orientable) Lorentzian metrics is essentially a question about the underlying topology.
Next, suppose $(M,g)$ is a Lorentzian manifold, and let $M_{\tau}$ be the set of all time cones in all the tangent spaces; then there is a canonical projection map $\pi:M_{\tau}\to M$, and it is possible to provide $M_{\tau}$ a topology and smooth structure such that $\pi$ becomes a smooth covering map, which is 2-1 (because there are exactly two time cones for each point). Then, $\pi$ is in particular a local diffeomorphism, so the pullback $g_{\tau}:=\pi^*g$ is a Lorentzian metric and with this, $\pi: (M_{\tau},g_{\tau})\to (M,g)$ becomes a Lorentzian covering called the time-orientation covering of $(M,g)$. Now, we have the relatively easy theorem:
Theorem (O’Neill, Chapter 7, Lemma 17).
Let $(M,g)$ be a Lorentzian manifold. Then,
- $(M_{\tau},g_{\tau})$ is time-orientable.
- $(M,g)$ is time-orientable if and only if $\pi:M_{\tau}\to M$ is trivializable.
Btw, you can also define a smooth action of $\Bbb{Z}_2$ (doesn’t matter if it’s right or left, since it’s Abelian) on $M_{\tau}$ where $-1$ acts by sending a timecone in $T_pM$ to the opposite timecone in $T_pM$. The orbit of a given point in $M_{\tau}$ (i.e of a given timecone) is simply the set of two timecones in a given tangent space, but this is precisely the fiber of $\pi$. Hence, by the universal property of quotient manifolds, we have $M\cong M_{\tau}/\Bbb{Z}_2$, and since the $\Bbb{Z}_2$ action on $M_{\tau}$ is free (obviously), it follows $(M_{\tau},\pi,M,\Bbb{Z}_2)$ is a principal bundle. Since $M_{\tau}$ is a principal bundle, the second statement is thus equivalent to $M_{\tau}$ admitting a global section. So, the theorem is saying that existence of a smooth timelike vector field in $(M,g)$ is equivalent to a smooth choice of timecones.
Also, there’s a relatively standard topological lemma (which you can google or try to prove yourself) that for a covering (say of manifolds) $\pi:X\to B$, if the base $B$ is simply connected, then $(X,\pi,B)$ is trivializable. Thus, we have the following corollary of (2) above:
Corollary.
Every simply-connected Lorentzian manifold is time-orientable.
Thus, for a given Lorentzian manifold, we have a sufficient condition on the topology which ensures time-orientability. But, I’m not aware of necessary and sufficient conditions (not that I’ve looked too hard).
Finally, it should also be noted that not every Lorentzian manifold is time-orientable, and indeed, once we consider manifolds which are not simply-connected, there are examples of Lorentzian metrics on the same manifold which are time-orientable, and also ones which are not time-orientable. In fact, even if we assume that the manifold is orientable (a topological assumption as you say), that still doesn’t really help us with time-orientability. So, time-orientability of a given Lorentzian metric is a combination of both topological and metric properties.
First let’s record the following easy lemmas (which are of inherent independent interest)
Lemma 1.
A product of smooth manifolds is orientable if and only if each factor is orientable.
By induction, it suffices to prove the case of two factors; call the manifolds $X,Y$, and let $m:=\dim Y$. If $X,Y$ are orientable, then letting $\mu_X,\mu_Y$ be respective volume forms, we have that $\mu:=\pi_X^*\mu_X\wedge\pi_Y^*\mu_Y$ will be a volume form. Conversely, suppose $X\times Y$ is orientable. Letting $Y_0$ be an open subset of $Y$ which is diffeomorphic to $\Bbb{R}^m$, we see that the restriction of any volume form on $X\times Y$ to $X\times Y_0$ is still a volume form. Hence, $X\times\Bbb{R}^m$ is orientable; let $\mu$ be a volume form here. Now, define $\mu_X:=\iota^*\left(\mu(\cdots, \eta_1,\dots, \eta_m)\right)$, where $\iota:X\to X\times\{0\}\subset X\times\Bbb{R}^m$ is the canonical embedding and for each $i\in\{1,\dots, m\}$, $\eta_i$ is the canonical lift of $\frac{\partial}{\partial y^i}$ on $\Bbb{R}^m$ to $X\times \Bbb{R}^m$. Then, $\mu_X$ is a volume-form on $X$, so $X$ is orientable. Reversing the roles of $X,Y$ in the proof above shows $Y$ is also orientable.
Lemma 2.
Let $(B,g_B)$ be a Lorentzian manifold, $(R,g_R)$ a Riemannian manifold, $\phi:B\to (0,\infty)$ any smooth function, and consider the warped Lorentzian manifold $M=B\times_{\phi}R$ (i.e $M=B\times R$ as a manifold, with metric $g=g_B+\phi^2 g_R$, or more precisely $g=\pi_B^*g_B+(\phi\circ \pi_B)^2 \pi_R^*g_R$). Then, $(B,g_B)$ is time-orientable if and only if $(M,g)$ is.
To prove the $(\implies)$ direction, take any timelike vector field $T$ on $B$, and define a vector field $\tau$ on $M=B\times R$ by
\begin{align}
\tau(b,r):=(T(b),0)\in T_bB\times T_rR\cong T_{(b,r)}(B\times R).
\end{align}
Then, obviously (suppressing the projections in the notation) we have $g(\tau,\tau)=g_B(T,T)<0$, so $\tau$ is timelike on $M$. Conversely, given any timelike vector field $\tau$ on $M$, we can separate it out into components $\tau(b,r)=(\tau_B(b,r),\tau_R(b,r))\in T_bB\times T_rR$ with $\tau_B,\tau_R$ being smooth vector fields on $M$. Observe that
\begin{align}
g_B(\tau_B,\tau_B)&=g(\tau,\tau)-\phi^2g_R(\tau_R,\tau_R)\leq g(\tau,\tau)<0,
\end{align}
where we have used that $g_R$ is Riemannian. So, if we fix $r\in R$ and define $T(b)=\tau_B(b,r)$, then the above calculation shows we have a smooth timelike vector field on $B$.
With these lemmas, it suffices to find 2-dimensional counterexamples, because to get one of higher dimension $n> 2$, we just take a product with an $(n-2)$-dimensional orientable Riemannian manifold (e.g $\Bbb{R}^{n-2}$ with usual positive-definite metric and orientation).
Counterexamples.
The open Mobius band $M$ (which is not an orientable manifold) admits a time-orientable Lorentzian metric, and also one which is not time-orientable.
The cylinder $S^1\times\Bbb{R}$ (which is an orientable manifold) admits a time-orientable Lorentzian metric, and also one which is not time-orientable.
Details for the Mobius band.
We claim the open Mobius band $M$ admits a time-orientable Lorentzian metric, and also one which is not time-orientable. Here are the details:
Let $\Bbb{Z}$ act on $\Bbb{R}^2$ by $n\cdot (x,y):=(x+n,(-1)^ny)$. This is a smooth, free and proper action (exercise), so by the orbit manifold theorem (see Lee’s Introduction to Smooth Manifolds, Theorem 21.10) the quotient $M:= \Bbb{R}^2/\Bbb{Z}$ equipped with the quotient topology carries a unique smooth structure such that the projection $\pi:\Bbb{R}^2\to M$ is a smooth submersion; this is one of the many possible definitions of the Mobius band (if you have some other definition, you can easily show it’s equivalent to this one). Now, let $X=\frac{\partial}{\partial x}$ on $\Bbb{R}^2$; this vector field is in fact $\Bbb{Z}$-equivariant as a map $X:\Bbb{R}^2\to T\Bbb{R}^2$ (relative to the given action on $\Bbb{R}^2$ and the induced tangent action on the tangent bundle), which implies that it descends uniquely to a smooth vector field $\widetilde{X}:M\to TM$ which is $\pi$-related to $X$. Intuitively, since the $x$-direction is simply being wrapped around, rather than being “twisted”, it follows that the $X$ vector field which points along the $x$-axis will have no problems descending to the quotient.
Now, consider the following Lorentzian metrics on $\Bbb{R}^2$: $g_1=-dx^2+dy^2$ and $g_2=dx^2-dy^2=-g_1$. Relative to both of these, the above action of $\Bbb{Z}$ on $\Bbb{R}^2$ is by isometries; and the fibers are discrete, so in particular semi-Riemannian manifolds in their own right, and so $g_1,g_2$ descend to Lorentzian metrics $\widetilde{g_1},\widetilde{g_2}$ on the quotient $M$, such that $g_i=\pi^*\widetilde{g_i}$ (Lee’s Introduction to Riemannian Manifolds, Corollary 2.29 covers the Riemannian case, but with minor modifications you can convince yourself the semi-Riemannian case holds as well). It is easily verified that
\begin{align}
\pi^*(\widetilde{g_i}(\widetilde{X},\widetilde{X}))=g_i(X,X)=(-1)^i.
\end{align}
So, for $i=1$, we see (by surjectivity of $\pi$) that $\widetilde{X}$ is a smooth timelike vector field on $(M,\widetilde{g}_1)$, and hence this Lorentzian metric on the Mobius band is time-orientable.
However, for $i=2$, the vector field $\widetilde{X}$ is spacelike, so supposing for the sake of contradiction there existed a smooth timelike vector field $T$ on $M$, it obviously can’t be a multiple of $\widetilde{X}$, and so $\{T,\widetilde{X}\}$ forms a smooth global frame for the tangent bundle. This implies $M$ is orientable as a manifold (since the bivector field $T\wedge\widetilde{X}$ is nowhere-vanishing, and thus by a musical isomorphism, we get a nowhere-vanishing 2-form on $M$), but this is absurd since we know the Mobius band is not an orientable manifold. Therefore, $(M,g_2)$ is not time-orientable.
For the sake of completeness, here’s a proof that $M$ is not an orientable manifold. Supposing it is, let $\mu$ be a volume form on $M$. Then $\pi^*\mu$ is a volume form on $\Bbb{R}^2$ since $\Bbb{Z}$ being discrete implies $\pi$ is a local diffeomorphism. Hence can be written as $f(x,y)\,dx\wedge dy$ for some smooth nowhere-vanishing function $f:\Bbb{R}^2\to\Bbb{R}$, which by the intermediate-value theorem must maintain aconstant sign. But now letting $\theta_n(x,y)=(x+n,(-1)^ny)$ be the action map of the integer $n$, we easily see by pulling back by $\theta_n$ that
\begin{align}
f\,dx\wedge dy&=\pi^*\mu=(\pi\circ\theta_n)^*\mu=\theta_n^*\pi^*\mu=\theta_n^*(f\,dx\wedge dy)=(-1)^n(f\circ \theta_n)\,dx\wedge dy,
\end{align}
which implies $f\circ \theta_n=(-1)^nf$, contradicting that $f$ maintains a constant sign (and in fact it says that if we move along the $x$-axis by $n$ units with $n$ odd, then $f$ flips sign, which coincides with our intuitive understanding of the ‘twist’ in the Mobius strip as well).
Details for the Cylinder.
The cylinder $M=S^1\times\Bbb{R}$ is obviously orientable and time-orientable with the metric $g=d\theta^2-dt^2$ (i.e the product of the usual positive-definite round metric on $S^1$ with the negative-definite $(\Bbb{R},-dt^2)$) because $\frac{\partial}{\partial t}$ is well-defined and timelike for $g$, so provides a time-orientation.
To give an example of a non-time-orientable Lorentzian metric on the cylinder, consider the following picture: 
The picture on the left is from O’Neill Chapter 5, figure 5 (page 145), and the picture drawn on the right is an example of a vector field in the covering space $\Bbb{R}^2$ of the cylinder. Intuitively, the red vector field rotates clockwise as we right along the $x$-axis, i.e the timecones are themselves rotating; and if we arrange the speed of rotation correctly,
the timecones after moving $2\pi$ will be in the flipped order, so when we project to the quotient, we’ll get an inconsistency.
More precisely, on $\Bbb{R}^2$ with coordinates $(x,y)$, let us define the tensor field
\begin{align}
g&=\cos x\,dx^2-2\sin x\,dx\,dy-\cos x\,dy^2
= -(dx^2+dy^2)+ 2\left(\cos\frac{x}{2}\,dx-\sin\frac{x}{2}\,dy\right)^2,
\end{align}
where in the second equality I have used the half-angle formulae. Note that the first equality tells us $g$ is symmetric. Next, define the vector fields
\begin{align}
T=-\sin\left(\frac{x}{2}\right)\frac{\partial}{\partial x} -\cos\left(\frac{x}{2}\right)\frac{\partial}{\partial y},\quad\text{and}\quad
X=\cos\left(\frac{x}{2}\right)\frac{\partial}{\partial x} -\sin\left(\frac{x}{2}\right)\frac{\partial}{\partial y}.
\end{align}
Then, using the second expression for $g$, it is easily seen that $g(T,T)=-1$ and $g(X,X)=1$, and with the first expression $g(T,X)=0$, proving that $g$ has Lorentzian signature and $\{T,X\}$ provides a global $g$-orthonormal frame with $T$ timelike and $X$ spacelike. Btw, the way I got these formulas is I first tried to write down a simple ‘rotating’ vector field like the red arrows above, and this easily gave me the formula for $T$. Then, I started with the general anszatz $g=\alpha\,dx^2+2\beta\,dx\,dy+\gamma\,dy^2$ and tried to choose $\alpha,\beta,\gamma$ to ensure $g(T,T)=-1$, and with a little guess-work I got the above formula, and lastly it was easy to see that $X$ as defined above satisfies $g(X,X)=1$ and $g(T,X)=0$, proving that $g$ is indeed Lorentzian.
Now, consider the action of $\Bbb{Z}$ on $\Bbb{R}^2$ defined as $n\cdot (x,y)\equiv \theta_n(x,y):= (x+2\pi n, y)$. The quotient is the cylinder: $M:=\Bbb{R}^2/_{\theta}\Bbb{Z}\cong (\Bbb{R}/2\pi \Bbb{Z})\times\Bbb{R}\cong S^1\times\Bbb{R}$. Also, the action consists of isometries of $g$: $\theta_n^*g=g$ (using the basic trigonometric facts that $\sin\left(\frac{x}{2}+n\pi\right)=(-1)^n\sin\left(\frac{x}{2}\right)$, and likewise for $\cos$, but this sign disappears because it is squared). Hence, $g$ descends to a Lorentzian metric $\widetilde{g}$ on the cylinder (if by abuse of notation you use local coordinates $(x,y)$ on $S^1\times \Bbb{R}$ with $x$ defined only mod $2\pi$, then it has the same formula as above). Also, it is easy to verify that $\theta_n^*T=(-1)^nT$ (the LHS being pullback of a vector field).
Suppose for the sake of contradiction $(M,\widetilde{g})$ is time-orientable, and let $\widetilde{T}$ be a smooth timelike vector field. Since $\Bbb{Z}$ is discrete, the projection $p:\Bbb{R}^2\to M$ is a covering map, and in particular a local diffeomorphism, so we can define a unique pullback vector field $\tau:=\pi^*\widetilde{T}$ (more precisely, there’s a unique vector field $\tau$ on $\Bbb{R}^2$ which is $p$-related to $\widetilde{T}$). Now, it is easy to verify using $g=p^*\widetilde{g}$ and that $\tau$ is $p$-related to $\widetilde{T}$ to prove that
\begin{align}
g(\tau,\tau)=p^*\left(\widetilde{g}(\widetilde{T},\widetilde{T})\right)<0,
\end{align}
and so $\tau$ is timelike. Also, it is straightforward to verify (keeping in mind $p\circ \theta_n=p$) that
\begin{align}
\theta_n^*\left(g(T,\tau)\right)&=g(\theta_n^*T,\tau)=(-1)^ng(T,\tau).
\end{align}
Now, since both $T$ an $\tau$ are timelike vector fields it follows $g(T,\tau)$ is nowhere-vanishing hence maintains a constant sign by the intermediate-value theorem. However, this contradicts the above equation which tells us that when $n$ is odd, $g(T,\tau)$ has opposite sign at $(x,y)$ and $\theta_n(x,y)=(x+2\pi n,y)$. Therefore, $(M,\widetilde{g})$ is not time-orientable.
This example specifically is nice in hindsight because the proof really does mimic the intuitive picture, and the contradiction reached via the intermediate-value theorem is a precise way of saying that when we travel $2\pi$ along the $x$-axis, the timecone has flipped.
Some final words about intuition: notice that in both of the examples above, the lack of time-orientability comes from the timecones in the quotient flipping direction as you travel around in a ‘single’ loop (which is only possible due to non simply-connectedness). But a key difference is the reason for the flip. In the Mobius band case, our timecones in the covering space $\Bbb{R}^2$ were the usual ‘upright’ ones, but in the quotient Mobius band, they flipped because the Mobius band itself is defined with a twist. In the cylinder case, the timecones in the covering space $\Bbb{R}^2$ themselves were the ones rotating, and since the quotient is obtained without a twist, it follows that the flip in the covering space still persists in the quotient. In both cases, formalizing the occurrence of the flip is done via the intermediate-value theorem.
