Consider $T^*M$ with the canonical symplectic structure.
Let $H:T^*M\rightarrow \mathbb{R}$ be an hamiltonian function and $h:\mathbb{R}\rightarrow \mathbb{R}$ a smooth function. Let $\gamma(t)$ be such that $\dot \gamma(t)=X_{h\circ H}(\gamma(t))$, where $X_{h\circ H}$ is the Hamiltonian vector field of $h\circ H$. Is it true that the quantity $2h'(H(\gamma(t)))H(\gamma(t))$ will be constant ? I know that $h\circ H(\gamma(t))$ is constant by definition of the hamiltonian vector field and flow , but I am not sure about what happens to the previous one. I have tried to compute it but I didn't get anywhere useful. One thing it led me to is to wonder wether $\omega(X_{h\circ H}, X_h)=0$ or not ?
Any help is appreciated. Thanks in advance.
Your expectations are correct on any symplectic manifold $(X, \omega)$. Roughly speaking, the level sets of $h \circ H$ are "the same" as those of $H$ and as such, $X_{h \circ H}$ and $X_{H}$ are colinear at every point.
A more rigorous argument (which is however insufficient to establish the colinearity statement I just mentioned) is to prove that $\omega(X_{h \circ H}, X_H) = 0$, which is the same as showing that the Poisson bracket $\{h \circ H, H\}$ vanishes. But the map $\{ - , F \} : C^{\infty}(X) \to C^{\infty}(X)$ is a derivation, so it satisfies the chain rule $\{h \circ H, F\} = (h'\circ H) \cdot \{H, F\}$ (check this identity in a Darboux chart if you wish). Since $\{H, H\} =0$, we get indeed $\{h \circ H, H\} = 0$.
It follows that $H$ is constant along the integral curves of $h \circ H$, which in turn implies that $h' \circ H$ is constant along those curves. The product $(h' \circ H) \cdot H$ is therefore also constant along those curves.