I begin by giving my answer to your question, which will involve ad hoc elements. Afterwards, I reflect upon the question "how to compute $\omega = d\theta$ in a coordinate-free fashion?", as it appears to me to be a rather vague question and also a rather unfortunate requirement.
Let $p : T^*M \to M$ denote the canonical projection, $p' : T(T^*M) \to T^*M$ denote the canonical projection and $Tp = p_* : T(T^*M) \to TM$ denote the differential of $p$. Then for $X \in T(T^*M)$, $\theta(X) := \langle p'(X) , Tp(X) \rangle$ where $\langle - , - \rangle$ is the duality product between $T^*_mM$ and $T_mM$ for any $m \in M$.
Given any smooth function $f : N \to Q$ between manifolds and given a $p$-form $\alpha$ on $Q$, it is well-known that $f^*(d\alpha) = d(f^*\alpha)$. We are going to apply this fact to different choices of $f : N \to T^*M$ with $\alpha = \theta$.
Let's first prove that the fibers of $p$ are isotropic submanifolds of the 2-form $\omega = d\theta$. Given such a fiber $T^*_m M$, there is the inclusions $\iota : T^*_mM \subset T^*M$, hence $\iota^*\omega = d(\iota^*\theta)$. But for $Y \in T(T^*_mM)$, $$(\iota^*\theta)(Y) = \theta(\iota_* Y) = \langle p'(\iota_*Y), p_*\iota_* Y \rangle = \langle p'(\iota_*Y), 0 \rangle = 0,$$ hence $d(\iota^*\theta) = 0$.
Given a nonzero $X \in T_mM$, there exists $\beta \in T^*_mM$ such that $\beta(X) = -1$ Observe that $\beta$ can be extended into a 1-form defined in a neighborhood of $m$ (and in fact on the whole of $M$); this 1-form can be identified with a section $s_{\beta} : M \to T^*M$ of $p$. Given $\gamma \in T^*_mM$, there is an element of $T_{\gamma}T^*_mM$ which is rather canonically identified with $\beta$, which we will denote $Y$.
Denoting $x,y$ the coordinates on $\mathbb{R}^2$, consider an immersion (this exists) $f : \mathbb{R}^2 \to T^*M$ with the following properties: $f(0,0) = \gamma$, $f_{*(0,0)}(\partial_y) = Y$ and $f_{*(0,0)}(\partial_x) = X'$ is a lift of $X$ i.e. $p_*X' = X$, which we extend in a neighborhood. Observe that $0 = f_* 0 = f_*[\partial_x, \partial_y] = [f_*\partial_x, f_* \partial_y]$. Then morally we compute
$$ \begin{align}
(f^*\omega)(\partial_x,\partial_y) &= \omega( X', Y) = X'(\theta(Y)) - Y(\theta(X')) - \theta([X', Y]) \\
&= X'(0) - Y(\gamma(X)) - \theta(0) = - \frac{d}{dt}[(\gamma + t \beta)(X)] = -\beta(X) = 1 \, .
\end{align} $$
This seems to prove that $\omega$ is nondegenerate at every point $\gamma \in T^*M$, and thus that it is therefore a symplectic form.
But this computation is not correct, as we need to extend $Y$ and $X'$ to be defined in a neighborhood of $\gamma$ in order to differentiate as we did. It is however sufficient for this computation that $f_*(\partial_y)$ be everywhere tangent to the fibers of $p$ (ideally chosen so that $f(0,t) = \gamma + t \beta$) and that, along the fiber $T^*_mM$, $f_{*}(\partial_x)$ be lifts of $X$. I leave to the reader the task of proving that such a $f$ exists...
The passage $Y(\gamma(x)) = \frac{d}{dt}[(\gamma + t \beta)(X)]$ might be harder to understand, except perhaps when $\gamma = 0$; it is due to the fact that $T^*M$ being a vector bundle, there is a canonical connection along each fibers of $p$. Nevertheless, I will rapidly sketch how we can reduce to the case $\gamma = 0$.
We now notice a following fundamental property of $\theta$. Given a 1-form $\beta$ on $M$, we can identify it with a section $s_{\beta} : M \to T^*M$ to $p$. For $X \in TM$, we compute $$ (s_{\beta}^*\theta)(X) = \theta(s_{\beta \, *}X) = \langle p'(s_{\beta \, *}X) , p_*s_{\beta \, *}X \rangle = \langle \beta , X \rangle = \beta(X) , $$
hence $s_{\beta}^*\theta = \beta$. Incidentally, $s_{\beta}^*\omega = d\beta$ and taking $\beta = 0$, we see that the 0-section is a Lagrangian submanifold for $\omega$.
Using this, we can show that whenever $\beta$ is a closed 1-form, the map $\Psi_{\beta} : T^*M \to T^*M : \gamma \mapsto \gamma + s_{\beta}(p(\gamma))$ is a (pre)symplectic map in the sense that it preserves $\theta$ and hence $\omega$ (this is not so easy to establish in a coordinate-free approach). Hence, taking $\beta$ to be a (closed) extension of $-\gamma$, the use of this map allows to compute $\omega$ only along the 0-section.
I now reflect upon the question "how to compute $\omega = d\theta$ in a coordinate-free fashion?"
First, by the very definition of a manifold, one could argue that there is no "coordinate-free" approach to differential geometry/topology, but at best a "non-manifestly coordinate-based" approach. Indeed, many of the arguments one could come with rely on constructions which are ultimately established by working in coordinates. When facing a new problem, it is unclear whether it could easily be solved using only already established constructions; perhaps a new construction would be more appropriate, and the way to achieve this construction might involve coordinates explicitly.
Secondly, what does "compute" mean? In the present context, my take on this question would be to describe $\omega$ in somewhat more explicit and intelligible terms. On the one hand, 'explicit' and 'intelligible' are vague qualificatives; the sought-after description should probably serve some definite purpose, for instance be fit for subsequent inquiry. On the other hand, the 'more intelligible terms' are probably ad hoc structures, objects with which we are more acquainted, such as coordinates; it seems arbitrary to snob the use of coordinates in favor of other referents. In that sense, the usual coordinate description of $\omega$ recalled in Camillo Arosemena-Serrato's answer is relevant, as it is intelligible and useful for further studies; incidentally and most interestingly, it shows that the canonical coordinates on $T^*M$ induced by any coordinates system on $M$ are Darboux coordinates. (By the way, this computation of $\omega$ is not independent of the coordinates on $T^*M$...)
Regarding the use of the formula $d\alpha(X,Y) = X(\alpha(Y)) - Y(\alpha(X)) - \alpha([X,Y])$ to $\alpha = \theta$, it is not a problem that $\theta$ be defined pointwise; any explicit description of a function on a set is a pointwise description, which does not prevent us from performing calculus on the function.
It is difficult to evaluate $d \alpha$ in a coordinate-free way, for it requires us to compare the values of $\alpha$ in neighboring points, which is difficult without some identification of cotangent spaces at neighboring points. Such an identification is provided by a coordinates system; it is also provided by a connection (but the existence of connection is best established by the use of coordinates...). In any case, the identification is an ad hoc structure to the problem.
The formula however provides a coordinate-free translation of the problem. Instead of having to compute $d\alpha$, then evaluate it at a point $p$ on vectors $X,Y \in T_pQ$, one rather extends $X$ and $Y$ to vector fields near $p$ (the extensions are ultimately irrelevant ad hoc structures) and compute the righthand side of the equation. In order for this technique to be manageable, one would be best to choose extensions $X$ and $Y$ suited to the problem (i.e. for the specific $Q$ and $\alpha$). In the situation at hand, one sees that the extensions provided by the coordinate vector fields $\partial/\partial x^j$ and $\partial/\partial \xi_k$ are very well suited to the computation of $d\theta$ through the righthand side of the equation.
