I read the picture below, but I don't know how to get the equation above red line. Whether by using $T_{\xi_x}(T^*M)\cong T^*_xM$ ?But which isomorphism should be choice ? Then , how to check the non-degenerate of $\omega$ ?
Picture below is from this web.
$x$ is an element of $M$, $TM$ the tangent bundle of $M$.$T_xM$ is the fiber of the projection $p:TM\rightarrow M$, $T^*M$ is the contangent bundle. If $\pi:T^*M\rightarrow M$ is the projection, the fiber $T^*M_x$ is the dual of $T_xM$. Thus an element $\epsilon_x\in T^*M_x$ is a linear form $\epsilon_x:T^*M_x\rightarrow R$.
The differential of $\pi$ is a morphism $d\pi:T(T^*M)\rightarrow TM$. Thus if $\epsilon_x\in T^*M_x, v_{\epsilon_x}\in T_{\epsilon_x}(T^*M), d\pi_{\epsilon_x}(v_{\epsilon_x})\in T_xM$, so since $\epsilon_x:TM_x\rightarrow R$, we can define the Liouville form by: $\epsilon_x(d\pi_{\epsilon_x}(v_{\epsilon_x}))$.
To check that $\omega$ is not degenerated, use local coordinates: Let $(U_i)_{i\in I}$ be a trivialisation of $TM$, $TU_i=U_i\times R^n$, $T^*U_i=U_i\times {R^n}^*$. An element $\epsilon_x\in {T^*U_i}_x$ is of the form $(x,e_x), e_x:R^n\rightarrow R$.$T({T^*U_i})=TU_i\times {R^n}^*$, an element of $T({T^*U_i}_{\epsilon_x})$ is of the form $v_x=(x,u_x,w_x)$ $d\pi(v_x)=(x,u_x)$. This implies that $\epsilon_x(d\pi(v_x))=e_x(u_x)$.
If you take a basis $(e_1,...,e_n)$ of $R^n$ and $(e_1^*,..,e_n^*)$ its dual which define trivialisation of $TU_i$ and $T^*U_i$ $(x,e_i)\in T_xU_i$, $(x,e_i^*)\in T^*_xU_i$, you have $(x,e_j^*)(d\pi(x,e_i,w_x)=e_j^*(e_i)$.