Consider a symplectic form $\omega_x(\xi,\nu)=\langle x,\xi\times \nu\rangle$ on $S^2$ where $x\in S^2$ and $\xi,\nu\in T_x S^2$ and a parametrization $\phi:U\to S^2$ where $U=(0,2\pi)\times(-1,1)$ and $\phi(\theta,x_3)=(\sqrt{1-x_3^2}\cos(\theta),\sqrt{1-x_3^2}\sin(\theta),x_3)$.
I want to show that $\phi^*\omega_x=\omega_{st}$ where $\omega_{st}=d\theta\wedge dx_3$. My logic is the following:
It's enough to show that $\phi^*\omega_x(v,w)=\omega_{st}(v,w)$ for all $v,w\in T_{p}U$ where $\phi(p)=x$. However, since $T_pU$ is spanned by two vectors $\partial_{\theta}$ and $\partial_{x_3}$, then it's enough to show that $$\phi^*\omega_x(\partial_{\theta},\partial_{x_3})=\omega_{st}(\partial_{\theta},\partial_{x_3})=1$$ where $\phi^*\omega_x(\partial_{\theta},\partial_{x_3})=\omega_x(d\phi_p(\partial_{\theta}),d\phi_p(\partial_{x_3})).$ By direct computation i.e. finding $d\phi_p$ and computing $\omega_x(d\phi_p(\partial_{\theta}),d\phi_p(\partial_{x_3}))$ explicitly by using a definition, we can check that $\phi^*\omega_x(\partial_{\theta},\partial_{x_3})=1$.
Therefore, $\phi^*\omega_x=\omega_{st}$ as they agree on the basis vectors.
I just want to check if this proof makes sense. Thank you!
Yes that's correct (assuming that you are checking $\phi^*\omega_x(\partial_{\theta},\partial_{x_3})=\omega_{st}(\partial_{\theta},\partial_{x_3})$ for all $p$). Note that the above argument works for any top form (not only for surfaces), since $\wedge^n \mathbb (R^n)^*$ is one dimensional, then two top forms $\omega_1, \omega_2 \in \wedge ^n\mathbb R^n $ are the same if and only if $$ \omega_1 (e_1, \cdots, e_n ) = \omega_2(e_1, \cdots, e_n)$$ for any fixed basis of $\mathbb R^n$.