Prove that there is a bundle isomorphism $\Phi : TM \oplus T^*M \to T(T^*M)$ which identifies the summand $T^*M$ with the vertical vectors.
If $\omega_{can}$ is the canonical symplectic form on the cotangent bundle $T^*M$, then prove further that one can choose $\Phi$ such that the composite $d\pi \circ \Phi$ restricts to the identity on the summand $TM$, where $\pi : T^*M\to M$ is the natural projection, and that $\Phi^*\omega_{can}=\Omega$ where $\Omega$ is defined by $\Big((v\oplus v^*),(w\oplus w^*)\Big) \mapsto w^*(v)-v^*(w)$.
Let $(U, x^i)$ is a local coordinate on $M$ near a point $p$, and $(T^*_UM, x^i, y^i)$ is a local coordinate on $T^*M$. Then the canonical form is given by $\omega_{can}= \sum_{i=1}^n dx^i \wedge dy^i$. My idea is to define $$ \Phi^{-1}: \frac{\partial}{\partial x^i} \mapsto \frac{\partial}{\partial x^i}, ~\frac{\partial}{\partial y^i} \mapsto dx^i $$
I try to construct $\Phi$ in local coordinates and show it is independent of the choice of coordinates. But this kind of argument seems ugly and requires lots of computation. If possible, I prefer intrinsic answers, as this problem itself seems elegant and concise.