I feel like this result is in a way obvious, since we can just apply the chain rule several times:
\begin{aligned}
\frac{d}{dt}(f \circ \phi^{-1} \circ \phi \circ \gamma)(0)
&= D_{f\circ\phi^{-1}\circ \phi}(\gamma(0)) \cdot \gamma' (0) \newline
&= D_{f\circ(\phi^{-1}\circ \phi)}(p) \cdot \gamma' (0) \newline
&=D_f(\phi^{-1}(\phi(p))) \cdot D_{\phi^{-1} \circ {\phi}} (p) \cdot \gamma'(0) \newline
&= D_f(p) \cdot D_{id}(p) \cdot \gamma'(0)
\end{aligned}
where $D$ denotes the Jacobian of the respective function.
We can do the same for the expression $\frac{d}{dt}(f \circ \psi^{-1} \circ \psi \circ \gamma)(0)$ and derive the same result.
Is that really it? Did I use somewhere that $f$ is a real valued function? Did I use that $\phi$ and $\psi$ are charts (except for the fact that the expression is well defined since the respective inverse maps exist)?