Misunderstanding of notation of tangent maps Im getting mad with the notation related to tangent bundles and so. Im following the book Analysis II of Amann and Escher, where the main discussed issue appear in page 270.

Some definitions: let $M\subset\Bbb R^n$ a submanifold of dimension $m$, then $$\begin{align}T_pM:=\{&(v)_p\in T_p\Bbb R^n:\\&\quad\exists\epsilon>0\exists\gamma\in C^1((-\epsilon,\epsilon),\Bbb R^n)\text{ such that }{\rm im}(\gamma)\subset M,\gamma(0)=p,\dot\gamma(0)=v\}\tag1\end{align}$$
where $T_p\Bbb R^n:=\{(p,v):v\in\Bbb R^n\}$ and $(v)_p:=(p,v)$. And for some differentiable $f:M\to Y$ is defined
$$T_pf:T_pM\to T_{f(p)}Y,\quad (p,v)\mapsto (f(p),\partial f(p)v)\tag2$$
Also is described a chain rule for tangent maps that reads
$$T_p(f\circ g)=T_{g(p)}f\circ T_p g\tag3$$
Suppose $\varphi$ is a chart of $M$ around $p$, thus is defined as in $(2)$, that is
$$T_p\varphi:T_pM\to T_{\varphi(p)}V$$
for some $V\subset\Bbb R^m$. 

My confusion starts from the representation of $(T_p\varphi)v$ (for some $v\in T_pM$). It is written as
$$(T_p\varphi)v=\sum_{k=1}^m v_ke_k\tag4$$
where it is not clear if $e_1,e_2,\ldots,e_m$ is a basis of $\Bbb R^m$ or a basis of $T_{\varphi(p)}\Bbb R^m$ (I know that both spaces are essentially the same thing anyway). 
The definition of differential of a map $f\in C^1(M,\Bbb R)$ is defined as
$$d_p f:={\rm pr_2}\circ T_p f\tag5$$
where ${\rm pr_2}$ is the canonical projection of the second coordinate. But here again appear my confusion with the notation of $(T_p\varphi)v$ when is stated that
$$(d_p f)v=\partial (f\circ\varphi^{-1})(x_0)(T_p\varphi)v\tag6$$
where $x_0:=\varphi(p)$. From all the above definitions I can follow the algebraic manipulations of $(6)$ but anyway it seems meaningless, observe that $(T_p\varphi)v\in T_{\varphi(p)}V\subset T_{\varphi(p)}\Bbb R^m$, however the domain of $\partial (f\circ \varphi^{-1})(x_0)$ is not $T_{\varphi(p)}V$!!!
I must assume that here the authors are using some trivial isomorphism between $T_{\varphi(p)}V$ and $V$?
 A: Since $T_p\mathbb{R}^n$ is canonically isomorphic to $\mathbb{R}^n$ via the map
$$\begin{align}
T_p\mathbb{R}^n &\to \mathbb{R}^n \\
(p,v) &\mapsto v
\end{align}$$
most authors do not bother distinguishing them. In fact, some even define $T_p\mathbb{R}^n$ to be equal to $\mathbb{R}^n$. Similarly, if $V$ is an open subset of $\mathbb{R}^n$, then the inclusion map $V \hookrightarrow \mathbb{R}^n$ induces an isomorphism $T_pV \cong T_p\mathbb{R}^n$ and hence an isomorphism $T_pV \cong \mathbb{R}^n$ for any $p \in V$. Thus $T_pV$ and $\mathbb{R}^n$ can be (and in practice, are) treated as if they were equal.

it is not clear if $e_1,e_2,…,e_m$ is a basis of $\mathbb{R}^m$ or a basis of $T_{\varphi(p)}\mathbb{R}^m$ 

It is a basis of $T_{\varphi(p)}V$, which is canonically identified with $T_{\varphi(p)}\mathbb{R}^n \cong \mathbb{R}^n$.

however the domain of $\partial(f\circ\varphi^{−1})(x_0)$ is not $T_{φ(p)}V$

Again, this is not a problem because $T_{\varphi(p)}V$ is, for all intents and purposes, equal to $\mathbb{R}^n$.
