Suppose $F: N \to M$ is $C^\infty$ at $p \in N$. Show that if $(U', \varphi')$ is any chart containing $p$ in the atlas of $N$ and $(V', \psi')$ is any chart containing $F(p)$ in the atlas of $M$, then $\psi' \circ F \circ (\varphi')^{-1}$ is $C^\infty$ at $\varphi'(p)$.
Since $\varphi'$ is a homeomorphism we have that $(\varphi')^{-1}(\varphi'(p))=p$ so $$\psi(F((\varphi')^{-1}(\varphi'(p))) = \psi(F(p))$$ but since $F(p)$ is in some atlas $(V', \psi')$ isn't this $\psi(F(p))$ $C^\infty$ by definition?
By definition of smoothness of $F$. There exists charts $(U,\phi)$ and $(V,\psi)$ such that you have $(\psi \circ F\circ\phi^{-1})$ is $C^{\infty}$.
Hence for any other pair $(U',\phi')$ and $(V',\psi')$ , you have:-
$\psi' \circ F \circ(\phi')^{-1}=(\psi'\circ\psi^{-1})\circ(\psi \circ F\circ\phi^{-1})\circ(\phi \circ (\phi')^{-1})$
Use the smooth compatibility the charts .
1.$(\psi'\circ\psi^{-1})$ is smooth by definition of smooth compatibility of charts
2.$(\psi \circ F\circ\phi^{-1})$ is smooth
And 3.$(\phi \circ (\phi')^{-1})$ is again smooth by definition of smooth compatibility of charts.
So the composition of these is also smooth. Hence you have your result.