How exactly does it make sense to differentiate a function whose input is a point on a manifold? In this lecture, prof Frederic Scheuller introduces the concept of the tangent vector space defined the following way by introducing the concept of velocity at a point:

Let $(M,\theta,A)$ be a smooth manifold, and a curve $\gamma: R \to M$ at least $C^1$(One differentiable and derivative is continous). Suppose $\gamma(\lambda_o)=p$, then the velocity at $p$ is defined as:
$v_{\gamma,p} : C^{\infty} (M) \to R$
Where $C^{\infty}(M) := \{ f: M \to R |  f   \text{ is a smooth function} \}$ equipped with an addition of the form $$(f+g)(p)= f(p) + g(p)$$
and s multiplication of the form $$( \lambda \cdot g)(p) = \lambda g(p)$$  where the addition and scaling is in the same sense as operations we define on the real numbers
From 1:23 of the video , time stamped link

The above is fine and more or less intuitive, but what bothers me is what happens at 52:51, where he takes the derivative of the following object:
$$[ \partial_i( f \circ x^{-1})] x (p)$$
I am not sure what the object of $\partial$ is precisely meant to mean but I can say that (this is one of my doubts) :
$p$ is the geometric point on the manifold
$x$ is the function which takes us from the manifold to the point on the chart corresponding to it
$x^{-1}$ is the function which inverts the point on the chart to the point on the manifold
$f$ is a function from a point on the manifold to real number line
This is where I get confused, if $x^{-1}$ maps from points on the chart to geometric points on the manifolds and f takes that as an input, so what would it mean to take derivative of such a thing? The professor even points this out at around 55:40 but doesn't explain what exactly he has written.
Thanks in advance.
 A: Let $M$ be an $n$-dimensional manifold, $p \in M$ a point, $x: U \subset M \to x(U) \subset\mathbb{R}^n$ a chart around $p$ and $f: V \subset M \to \mathbb{R}$ a function such that $p \in V$ (with $V$ open in $M$). A priori it doesn't make any sense to say $f$ is smooth at $p$. But we can differentiate functions with Euclidean (open) domains and co-domains, and the coordinate representation of $f$ with respect to the chart $x$ fits that exact description! So you ask, what is the coordinate representation of $f$? Well, since $x$ is a chart, we can map points of $x(U)$ to $M$ via $x^{-1}$ and look at which of those points belong to $V$ so we can apply $f$ to those points. Doing that, we get a function $\hat{f}: x(U \cap V) \subset \mathbb{R}^n \to \mathbb{R}$, and we do know how to differentiate such a function. So we shall say that $f$ is smooth at $p$ iff its coordinate representation $\hat{f} \doteq f \circ x^{-1}$ is smooth at $x(p)$. Since $x(p)$ is a point in $\mathbb{R}^n$, all of this makes perfect sense. In particular, when $f$ is smooth, we can talk about the partial derivative of $f$ at $p$ by defining them to be the partial derivatives of its coordinate representation relative to a chart (again, we do know very well how to take derivatives in the Euclidean case) evaluated at... well, at the only place where it makes sense to evaluate it: $x(p)$. Since $\hat{f}$ is a function from $\mathbb{R}^n$ to $\mathbb{R}$, there are $n$ such partial derivatives. Naturally, they are given by:
$$\frac{\partial f}{\partial x^i}(p) \doteq \frac{\partial (f \circ x^{-1})}{\partial r^i}(x(p)), \text{for each $1 \leq i \leq n$ }$$
where on the left side the partial derivative is with respect to the chart $x$ and on the right side, it's the ordinary Euclidean partial derivative (which is again a partial derivative with respect to a chart, just the most trivial one of them all). And of course, all of this generalizes very naturally to functions between general manifolds.
A: Very clearly $(\partial f\circ x^{-1})x(p)$ is not what was written on the board; I assume you just made a typo. What was written is $[\partial_i(f\circ x^{-1})](x(p))$. This is simply the $i^{th}$ partial derivative of the function $f\circ x^{-1}:x[U]\subset\Bbb{R}^d\to\Bbb{R}$ evaluated at the point $x(p)$. Now this is really just basic multivariable calculus at this point. If you want everything explicitly in terms of limits, we have
\begin{align}
\frac{\partial f}{\partial x^i}(p)&:= [\partial_i(f\circ x^{-1})](x(p))\\
&:=\frac{d}{ds}\bigg|_{s=0}(f\circ x^{-1})(x(p)+se_i)\\
&:=\lim_{s\to 0}\frac{(f\circ x^{-1})(x(p)+se_i) - (f\circ x^{-1})(x(p))}{s}\\
&=\lim_{s\to 0}\frac{(f\circ x^{-1})(x(p)+se_i)-f(p)}{s},
\end{align}
where $e_i=(0,\dots, \underbrace{1}_{\text{$i^{th}$ spot}},\dots, 0)\in\Bbb{R}^d$. The first equal sign is a definition for the symbol on the LHS in terms of usual partial derivatives. THe next equal sign is simply the definition of the $i^{th}$ partial derivative. The next equal sign is just the definition of a (single-variable) derivative in terms of a difference quotient. The last equal sign is just by definition of composition.
Here everything makes perfect sense. $f\circ x^{-1}$ is a function from an open subset $x[U]$ of $\Bbb{R}^d$ into $\Bbb{R}$. $x(p)$ is an element of $x[U]$. So, by openness, if $s\in\Bbb{R}$ is small enough, then the points $x(p)+se_i$ lie in $x[U]$. Thus, it makes sense to plug these into $f\circ x^{-1}$.
