Let $M$ be a manifold and denote by $T^*M$ its cotangent bundle. Let $(x,U)$ be a coordinate chart so that $x: U\to \mathbb{R}^{n}$. Let $p\in U$ and $v\in T^*_pM$, then we can write $v = v_i dx^i$ where $v_i=v(\partial_{x^i})$. More in general, a cotangent vector $w\in T^*M\vert_U$ is given by $$ w = w_i dx^i $$ where now $w_i:U\to \mathbb{R}$. This defines a chart on $T^*M\vert_U$ via $$ (x,y):T^*M\vert_U\to \mathbb{R}^{2n}\\ (p,v)\mapsto (x(p),v(\partial_{x^i})). $$ All this is well known and so is the fact that $$ \omega = dx\wedge dy = dx^i\wedge dy_i $$ defines a symplectic structure on $T^*M$.
So we may consider curves $\gamma,\sigma: \mathbb{R} \to T^*M$ going through $(p,v)\in T^*M\vert_U$ at time $0$ and evaluate $\omega$ at $\dot{\gamma}(0),\dot{\sigma}(0)$.
My question is: how do we compute $\omega_{(p,v)}(\dot{\gamma}(0),\dot{\sigma}(0))$ explicitly?
I consider an explicit example with given curves and charts. The general case follows from changing the notation.
Let $M=S^1$, $x^1$ the "projection left" of $S^1\cap\{u^1>0\}$ and $$ \gamma_0(t)=(\cos(at),\sin(at))\\ \sigma_0(t)=(\cos(bt),\sin(bt)) $$ for $a,b\in\mathbb{R}$. Then a possible curve in $T^*M$ is given by $$ \gamma(t) = (\gamma_0(t),f(\gamma_0(t))dx) \in \Gamma_{\gamma_0}(T^*M) $$ for some $f:M\to \mathbb{R}$ and we do similarly with $\sigma$ using a function $g:M\to\mathbb{R}$.
Let's now evaluate $\omega$. We first compute $$ dx^1\dot{\gamma}(0)= \frac{d}{dt}\vert_{t=0}x^1\gamma_0(t)=\frac{d}{dt}\vert_{t=0} \sin(at)=a $$ while $$ dy^1\dot{\gamma}(0)= \frac{d}{dt}\vert_{t=0}\gamma(t)\partial_{x^1}=\frac{d}{dt}\vert_{t=0} f\gamma_0(t)=(df) \dot{\gamma}_0(0). $$
So if for example $f:S^1\to \mathbb{R}$ sends $(u^1,u^2)\mapsto \alpha u^2$ and $g:S^1\to \mathbb{R}$ sends $(u^1,u^2)\mapsto \beta u^2$ we get $$ \omega(\dot{\gamma}(0),\dot{\sigma}(0)) = ab\beta-ba\alpha. $$