I don't know much about differential geometry but as I understand it for a map between manifolds $F : M \to N$ to be $C^1$ around $p$ means that there exists charts $(U,\phi)$ containing $p$ and $(V,\psi)$ containing $F(p)$ such that $\psi \circ F\circ\phi^{-1}$ is $C^1$. But I think I also saw something saying that $F$ is $C^1$ means that the function that maps a point $p$ to its differential (i.e. $p \mapsto dF_p$) is continuous. Is that correct? How can I show that these two notions are equivalent?
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Already, to talk about the continuity of $p \mapsto dF_p$ what do we take as topology on $T_p(N)$? Do we consider the standard topology on $\mathbb{R}^n$ and translate it to $T_pN \cong \mathbb{R}^n$ via the basis $\{d\phi_1,\ldots,d\phi_n \}$?
It is indeed, correct. To show these notions are equivalent, note that the continuity $dF:TM\to TN$ is a local matter, so that one can work with charts $U\subset M,\; V\subset N$, borrowing your notation. Let $x_1,...,x_m$ be coordinates in $U$ and $y_1,...,y_n$ coordinates in $V$ (your $\psi,\phi$ respectively). So $dF(x_1,...,x_m,v)=(F_1(x),...,F_n(x),dF_{x}(v))$, So you see that it is continuous if and only if $\psi\circ F\circ\psi^{-1}$ is continuous, and if the partial derivatives of $F$ in the $x$ coordinates are continuous functions of $x$, which is the same as $\psi\circ F\circ\phi^{-1}$ being $C^1$ in the "usual" sense.