I have been reading Differential Topology by Guillemin and Pollack recently and I came across the definition of derivative for a smooth map between two manifolds.
Let $f:X\rightarrow Y$ is a smooth map between manifolds. We can find local parameterizations $\phi$ and $\psi$ for $X$ and $Y$ respectively with $\phi(0)=x$ and $\psi(0)=y=f(x)$ such that the following commutative diagram holds.
$$\require{AMScd} \begin{CD} X @>f>>Y\\ @A\phi AA @A\psi AA\\ U @>h\ =\ \psi^{-1}\ \circ\ f\ \circ\ \ \phi >> V \end{CD}$$
Now we know that $d\phi_0,\ d\psi_0,$ and $dh_0$ exist since either the maps are local parameterizations or between open subsets of Euclidean spaces. Thus the following commutative diagram will also hold.
$$\require{AMScd} \begin{CD} T_x(X) @>df_x>>T_y(Y)\\ @Ad\phi_0 AA @Ad\psi_0 AA\\ \mathbb{R}^m @>dh_0 >> \mathbb{R^n} \end{CD}$$
So we can define the derivative of smooth $f:X\rightarrow Y$ as:
$$df_x=d\psi_0\ \circ\ dh_0\ \circ\ d\phi_0^{-1}$$
Now I am wondering that from the first commutative diagram, we can say $f=\psi \ \circ\ h \ \circ \phi^{-1}$.
Using that, we define $df_x$. Since $f$ is composition of local parameterizations and map between open subsets of euclidean spaces, we can try to apply chain rule to get:
$$df_x=d\psi_{h(\phi^{-1}(x))}\ \circ\ dh_{\phi^{-1}(x)}\ \circ\ d\phi_x^{-1} \\ =d\psi_0\ \circ\ dh_0\ \circ\ d\phi_x^{-1}$$
Now this is clearly different from the previous definition we got from the second commutative diagram.
So my query is, can the chain rule be applied for composition of local parameterizations and maps between subsets of euclidean spaces? If yes, then why do we get two different definitions for the derivative of a smooth map between two manifolds? My apologies if this is too trivial.
THANK YOU