The reference text is N.M.J. Woodhouse's Geometric Quantization.
Let $Q$ be a smooth manifold with coordinates $\{q_i\}$, and $M=T^*Q$ be its cotangent bundle with coordinates $(p_i, q_i)$ (so that $\{p_i\}$ will characterize a covector).
It has been stated that, for the 1-form $\theta = p_adq^a, \omega=d\theta$ is a symplectic form on $M$ ($\omega=dp^a\wedge dq_a-$note that I'm using the Einstein summation convention). So far, I'm on board with things.
But now, in order to show that $\theta$ is coordinate-independent, it has been characterized in the following alternative manner:
$\theta_{(p, q)}:=(X_{(p, q)}\mapsto p(\pi_*X_{(p, q)}))$, where $X\in\chi(M), \pi_*$ is the pushforward of the projection $(p, q)\mapsto q$.
I'm unable to figure out how these two are equivalent. I suppose that in particular, what needs to be shown is that $p_adq^a(p, q)(X)=p(\pi_*X)$. Any help on the matter?
Since I need to practice that myself I'd like to give it a go:
The vector field $X$ can be written as \begin{align} X&=u^a\partial_{p_a}+v^a\partial_{q_a}\,,\\[2mm] \end{align} with functions $u^a,v^a$ on $M$ (i.e. functions "of" $\{p_i,q_i\}$)
For any differentiable function $f$ on $M\,,$ we see from $\pi:(p,q)\mapsto q$ that \begin{align}\tag{1} (\pi_*X)(f)&:=X(f\circ\pi)=(v^a\partial_{q_a}f)\,. \end{align} To see this note that by definition of the projection $(f\circ\pi)(p,q)=f(q)\,.$ Therefore, the LHS of (1) does nothing else than applying $X$ to $f$ which depends only on $q_1,...,q_n\,.$
In short: (1) is \begin{align} \pi_*X&=v^a\partial_{q_a}\in TQ\,. \end{align} Therefore, \begin{align} \theta(\pi_*X)=(p_b\,dq^b)(v^a\partial_{q_a})&=p_av^a\,. \end{align} Since I don't have the book I assume that the authors identify $p_av^a$ with $p(v^a\partial_{q_a})\,.$