I'll try and put together an answer to adress the points mentioned.
First of all, orientability is a property of manifolds, not of general spaces. Given a manifold $M$ of dimension $n$, one can ask whether it is possible to coherently give local orientations around each point, which from the viewpoint of homology means a choice of generator $\mu_x \in H_n(M, M \setminus \{x\})$ such that for any open coordinate ball $U \subseteq M$ there is a class $\mu_U \in H_n(M, M \setminus U)$ mapping to $\mu_x$ for all $x \in U$ under the restriction. If it is, then we call $M$ orientable and the $\mu_x$ an orientation of/for $M$.
If $M$ is smooth, then one can alternatively define an orientation of $M$ as an orientation of its tangent bundle (i.e. a vector-space orientation of each fibre in a coherent way), to give a definition that's perhaps a little more intuitive.
An orientation on a manifold imposes very strong conditions on its co/homology. Assuming that $M$ is closed, for simplicity, it satisfies Poincaré Duality: There is an isomorphism $H_k(M) \overset{\cong}{\to} H^{n - k}(M)$ for all $k$ (and moreover this isomorphism is of a particularly useful nature that I won't go into here).
To see just how strong this is, it immediately drops out of the statement that $\mathbb{R}\mathrm{P}^2$ is not orientable: If it were, we would have $0 = H_2(\mathbb{R}\mathrm{P}^2) \cong H^0(\mathbb{R}\mathrm{P}^2) \cong \mathbb{Z}$ from duality, which is absurd. Invoking the stronger statement that $H_n(M) \cong \mathbb{Z}$ iff $M$ is orientable and 0 else, usually proved en route to the duality theorem, we see that $\mathbb{R}\mathrm{P}^3$, however, is orientable, since it has $H_3(\mathbb{R}\mathrm{P}^3) \cong \mathbb{Z}$.
This should be enough to show that existence of a torsion element in $H_*(M)$, in particular in $H_1(M)$ is not enough to say anything about orientability (and in this case we even have that $\mathbb{R}\mathrm{P}^2$ embeds into $\mathbb{R}\mathrm{P}^3$ in a way that induces an isomorphism on fundamental groups).
Rather, orientability is related to the fundamental group in a certain way: if $\pi_1(M)$ does not have an index 2 subgroup, then it is orientable (this relates to a 2-sheeted covering of $M$ one can construct from its local orientations, called the orientation cover, that's trivial iff $M$ is orientable). This for instance applies to show that manifolds with fundamental groups like $\mathbb{Z} / p$ for $p > 2$ prime are automatically orientable.
Finally, let me address the question of whether a non-orientable manifold always has non-trivial torsion in homology. I will again assume for simplicity's sake that $M$ is closed. To tackle this question, I will need to make an observation: All the discussion of orientations and duality above implicitly used $\mathbb{Z}$-coefficients, but there's really nothing stopping me from switching that out for any other nice ring; in particular for a field. Looking at the definition of orientation, if I take $\mathbb{F}_2$-coefficients then all the local groups $H_n(M, M \setminus \{x\}; \mathbb{F}_2)$ will be 1-dimensional $\mathbb{F}_2$-vector spaces and therefore contain a unique non-trivial element, so I have to define the $\mu_x$ to be that very element and get a unique orientation for free.
In other words, every manifold is $\mathbb{F}_2$-orientable, and thus has $\mathbb{F}_2$-Poincaré duality. I thus conclude that $H_n(M; \mathbb{F}_2) \cong H^0(M; \mathbb{F}_2) \cong \mathbb{F}_2$, but now I can apply the universal coefficient theorem to get
$$
\mathbb{F}_2 \cong H_n(M; \mathbb{F}_2) \cong H_n(M; \mathbb{Z}) \oplus \operatorname{Tor}(H_{n - 1}(M; \mathbb{Z}), \mathbb{F}_2) \cong \operatorname{Tor}(H_{n - 1}(M; \mathbb{Z}), \mathbb{F}_2)
$$
which implies that $H_{n - 1}(M; \mathbb{Z})$ has 2-torsion, so the answer to the question is "yes."
