A doubt about a proof of chain rule for smooth functions between smooth manifolds I'm reading Theorem 1.1 in this lecture notes.

Theorem 1.1 (Chain Rule for Manifolds). Suppose $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ are smooth maps of manifolds. Then:
$$
\mathrm{d}(g \circ f)_{x}=(\mathrm{d} g)_{f(x)} \circ(\mathrm{d} f)_{x}
$$
Proof. If $\varphi$ is a local parameterisation of $x, \psi$ is a local parameterisation of $y=f(x)$, and $\eta$ is a local parameterisation of $z=g(f(x))$, then this is evident from:
$$
g \circ f=\left[\eta \circ\left(\eta^{-1} \circ g \circ \psi\right) \circ \psi^{-1}\right] \circ\left[\psi \circ\left(\psi^{-1} \circ f \circ \varphi\right) \circ \varphi^{-1}\right]
$$
and the usual chain rule from multivariate analysis (as we can differentiate the RHS as before using the usual chain rule).


In the proof, $g \circ f = \Psi \circ \Phi$ with \begin{align}
\Psi &:= \eta \circ\left(\eta^{-1} \circ g \circ \psi\right) \circ \psi^{-1} \\
\Phi &:= \psi \circ\left(\psi^{-1} \circ f \circ \varphi\right) \circ \varphi^{-1}.
\end{align}
The composition of smooth maps is also smooth, so $\Psi, \Phi$ are smooth. However, $\operatorname{dom} (\Psi) = \operatorname{dom} (\psi^{-1})$ which is open in $Y$ but not necessarily open in its ambient Euclidean space. The same situation holds for $\Phi$. So $\Psi, \Phi$ are not differentiable in the usual sense, so we can not apply the chain rule on $\Psi \circ \Phi$. However, the author said

"...we can differentiate the RHS as before using the usual chain rule...".

Could you please elaborate on my confusion?
 A: In definition 1.8 of the lecture notes, it is said that if $f\colon X \to Y$ is smooth and if $\varphi \colon U \to X$ and $\psi \colon V \to Y$ are local parametrisations in neighbourhoods of $x$ and $f(x)$, then
$$
df_x = d\psi_{\psi^{-1}(f(x))} \circ d(\psi^{-1}\circ f \circ \varphi)_{\varphi^{-1}(x)} \circ \big[(d\varphi_{\varphi^{-1}(x)})^{-1} \big]
$$
where $d\varphi_{\varphi^{-1}(x)}$ is considered as an isomorphism from $\Bbb R^n \to T_xX$, that is, forgeting about the ambiant euclidean space $\Bbb R^N$ around $X$.
Same thing with $g$ at the point $f(x)$ gives
$$
dg_{f(x)} = d\eta_{\eta^{-1}(g(f(x)))} \circ d(\eta^{-1}\circ g \circ \psi)_{\psi^{-1}(f(x))} \circ \big[(d\psi_{\psi^{-1}(f(x))})^{-1} \big]
$$
where $d\psi_{\psi^{-1}(f(x))}$ is considered as an isomiorphism from $\Bbb R^{\tilde{n}} \to T_{f(x)}Y$, that is, forgeting about the ambiant euclidean space $\Bbb R^{\tilde{N}}$ around $Y$.
For $g\circ f$, it yields
$$
d(g\circ f)_{g(f(x))} = d\eta_{\eta^{-1}(g(f(x)))} \circ d(\eta^{-1}\circ (g\circ f) \circ \varphi)_{\varphi^{-1}(x)} \circ \big[(d\varphi_{\varphi^{-1}(x)})^{-1} \big]
$$
You are thus lead to show that
$$
d(\eta^{-1}\circ (g\circ f) \circ \varphi)_{\varphi^{-1}(x)}= \big(d(\eta^{-1}\circ g \circ \psi)_{\psi^{-1}(f(x))}\big)\circ\big( d(\psi^{-1}\circ f \circ \varphi)_{\varphi^{-1}(x)}\big)
$$
This is just the chain-rule applied to the composition of
$$
\psi^{-1}\circ f \circ \varphi \colon U \subset \Bbb R^n \to V \subset \Bbb R^{\tilde{n}}  
$$
and
$$
\eta^{-1}\circ g\circ \psi \colon V \subset \Bbb R^{\tilde{n}} \to W \subset \Bbb R^{\tilde{\tilde{n}}}
$$
which are smooth functions defined on and with range in open subsets of euclidean spaces.
By the way: there is an obvious typo in the author's sketch: the arrow of at bottom right corner should be $\eta^{-1}\circ g \circ \psi$, not $\eta^{-1}\circ f \circ \psi$.
