How do I make sense of the following definitions of tangent vectors? I am teaching myself differential geometry in manifolds. I am reading a book, which at the beginning sets some notation that might be important for my question so I report  it here just in case it is not standard
Some notation
"$u^i $" is the n-th natural coordinate (projection) $\mathbb{R}^n \to \mathbb{R}$, so that $u=(u^1,...,u^n)$ is the identity function of $\mathbb{R}^n$. For a chart $(U,x)$The n functions $x^1,...,x^n: U \to \mathbb{R}$, components of $x$ given by $x^i=u^i\circ x$ are coordinate (functions) in $U$ with respect to the chart. $(x^1,...,x^n)$ is a local coordinate system and $x^1,...,x^n$ are called local coordinates, at $p$, if $p$ is a point of $U$
Applied vectors as tangent vectors .
Let $p$ be a point of  $\mathbb{R}^n$.
The operator $D_k|_p : f \in C^\infty(\mathbb{R}^n) \to D_kf(p)=\frac{\partial f}{\partial u^k}(p) \in \mathbb{R}$ is a tangent vector to $\mathbb{R}^n$at  $p$...(1)
So that to each applied vector$ (p,\nu) \in \{p\}\times \mathbb{R}^n $ it is associated a tangent vector $D_{(p,\nu)} \in T_p\mathbb{R}^n$ defined as $D_{(p,\nu)}f=(d/dt)_0 (f(p+t\nu))$ , directional derivative of $f$ at $p$ with respect to $\nu$. This can be written as $D_{(p,\nu)}f= \nu^k D_k|_pf=\nu^k D_{(p,e_k)}f$
Tangent vectors associated to a local coordinate system
Let $p$ be a point of the n-manifold $M$, to each chart $(U,x)$, there are $n$ tangent vectors associated at $p$. The k-th one is
$\frac{\partial}{\partial x^k}|_p : f \in C^\infty(M) \to \frac{\partial f}{\partial x^k}(p)=D_k(f \circ x^{-1})(x(p))$...(2)
So regarding this two points my questions are:

*

*I am a bit confused about the use of $x$ vs $u$. $x^i$ are coordinates but at the same time the components of the homeomorphism $x$ In (1), if the coordinates are supposed to be $x^1,...,x^n$, why is the partial derivative taken with respect to the identity function components $\frac{\partial }{\partial u^k}$?


*Regarding (2), Is this $\frac{\partial f}{\partial x^k}(p)=D_k(f \circ x^{-1})(x(p))$ a definition or is it something that can be proven?. If I try to do the derivative of the composition at some point I have to derivate the inverse function "$x^{-1}$", and I am getting nowhere. Where does the "$\circ x^{-1}$" even come from?
 A: This is the point: to work on a manifold we use coordinate charts $(U,x)$ so that, in this chart, we work as in $\mathbb{R}^n$, WLOG we can assume that $x: U\to \mathbb{R}^n$. Differential geometry is constructed in a way such that partial derivatives represented by $\frac{\partial}{\partial u^k}$ in $\mathbb{R}^n$, in $U$ under the coordinate system $x$ are denoted by $\frac{\partial}{\partial x^k}$.
That is: $x$ act as a diffeomorphism between $U\subset M$ and $\mathbb{R}^n$, therefore if we have a differentiable function $f:M\to \mathbb{R}$, where $M$ is a manifold, then using the coordinate system $x$ in $U$ we define
$$
\frac{\partial}{\partial x^k}\bigg|_p f:=\frac{\partial}{\partial u^k}\bigg|_{x(p)}(f\circ x^{-1})
$$
That is: to differentiate $f$ we do it locally using coordinate charts, we use the calculus in $\mathbb{R}^n$ to define a local calculus in $M$. This is all. We use the Euclidean spaces to model a differential calculus in what we call differentiable manifolds: sets with an structure such that locally are diffeomorphic to an Euclidean space.
ADDITION: the $\circ x^{-1}$ is needed because we are transporting $f$ to $\mathbb{R}^n$, and we do it using the diffeomorphism $x$. Note that the domain of $f$ is $M$, not $\mathbb{R}^n$.
