In one differential geometry class I'm taking, a critical point $p$ of a function $f$ on a manifold $(M,\mathcal{A})$ is defined as a point where $\forall$ coordinate chart $(U,\phi_U)\in\mathcal{A}$, $Df\circ\phi_U^{-1}(\phi_U(p))=0$. However, I don't understand why do we want to define it this way. Why can't we define it simply as $Df(p)=0$? For example, it is possible that $D\phi_U^{-1}$ is the zero map at $\phi_U(p)$, but $Df(p)\neq 0$, and then $Df\circ\phi_U^{-1}(\phi_U(p))=Df(p)\circ D\phi_U^{-1}(\phi_U(p))=0$
Edit: Here’s an example showing that the definitions are not equivalent. Consider $S^2$ and the chart $(U,\phi_U)$, where $U=S^2-\{(0,0,1)\}$ and $\phi_U(x_1,x_2,x_3)=(\frac{x_1}{1-x_3},\frac{x_2}{1-x_3})$.$\phi^{−1}_U(u,v)=(\frac{2u}{u^2+v^2+1}, \frac{2v}{u^2+v^2+1}, \frac{u^2+v^2-1}{u^2+v^2+1}).$ Then for $f(x_1,x_2,x_3)=(x_3)^2$, $Df\circ\phi^{-1}_U$ is zero at $(0,0)$. This shows that $(0,0,−1)$ is a critical point of $f$ on $S^2$. However, $Df(0,0,−1)=(0,0,−2)\neq 0$
First of all, it is unclear if in the class you are taking they defined $Df$ for smooth functions $f$ on $M$. I assume they did and that they verified that for smooth maps of open subsets $f: R^n\to R^m$, $Df$ is given by the "usual" derivative from the vector calculus. I assume also that they proved in the class the Chain Rule for maps between differentiable manifolds: If $$ X \stackrel{g}{\longrightarrow} Y \stackrel{f}{\longrightarrow} Z $$ are smooth maps of smooth manifolds, then for $h=f\circ g$ one has $$ Dh=Df \circ Dg. $$ Furthermore, I assume that you verified that local charts $(\phi, U)$ on a smooth manifold are diffeomorphisms between their domains and ranges, in particular, $D\phi^{-1}_q$ is invertible for all $q\in \phi(U)$.
Now, the definition for a critical point of $f\in C^\infty(M)$, $$ Df_p={\bf 0} $$ is perfectly reasonable. (Here and below, ${\bf 0}$ stands for the zero map $T_pM\to {\mathbb R}$.) Let's verify that it is equivalent to the definition from your class:
Suppose that $(U, \phi)$ is a chart on $M$, $p\in U, q=\phi(p)$. Then for $g=\phi^{-1}$ and $h=f\circ g$ we have $$ Dh_q= Df_p \circ Dg_q. $$ Since $Dg_q$ is invertible, it follows that $$ Dh_q\circ (Dg_q)^{-1}= Df_p $$
The claim then is that $Dh_q$ if and only if $Df_p={\bf 0}$. But this is just a simple linear algebra statement: If $Df_p={\bf 0}$ then $Dh_q={\bf 0}\circ Dg_q={\bf 0}$. If $Dh_q={\bf 0}$ then $$ {\bf 0}\circ (Dg_q)^{-1}= Df_p \Rightarrow Df_p={\bf 0}. $$
Thus, the two definitions of critical points are equivalent.