Tangential Space of a differentiable manifold is always $\mathbb R^n$? Let $\mathcal M$ be a differential manifold with a point $p$. Let U be an open set, $p\in U$, on $\mathcal M$ and let $\phi,\psi:U\to \mathbb R^n$ be a charts on $\mathcal M$. I'm having diffculties arranging all the concepts of a differential manifold. The "manifold" of required spaces gives me a headache. At the moment im utterly confused with the construction of the tangent space at $p$.
I'll start describing my thoughts from the start: $\require{AMScd}$
\begin{CD}
 U   @>\phi>>  \mathbb R_\phi^n \\
@V \psi VV @.\\
\mathbb R_\psi^n 
\end{CD}
That's the open set U with charts $\psi,\phi$. I denoted $\mathbb R^n$ with the chart pointing to it, so we can make sure to differentiate between the different $\mathbb R^n$ -spaces we will encounter.
The tangent space of $\mathcal M$ or $U$ in $p$ is a $n$-dimensional $\mathbb R$-vector space, which has isomorpisms pointing to $T_{\phi(p)} \mathbb R^n$ and $T_{\psi(p)} \mathbb R^n$. 
Where $T_{\phi(p)} \mathbb R^n$ and $T_{\psi(p)} \mathbb R^n$ are vector spaces we're inducing by the choice of our bases $B_\psi , B_\phi$ depending on our charts as follows after the diagram:
\begin{CD}
T_p \mathcal M   @> \cong >>  T_{\phi(p)} \mathbb R^n \\
@V \cong  VV @.\\
T_ {\psi(p)} \mathbb R^n 
\end{CD}
$T_{\phi(p)} \mathbb R^n$ and $T_{\phi(p)} \mathbb R^n$ are induced by the vector space bases $B_\psi , B_\phi$ using isomorphisms from the images of our charts ($\phi(U)\subset \mathbb R_{\phi}^n, \psi(U)\subset\mathbb R_{\psi}^n)$ to the tangent space $T_{\phi(p)} \mathbb R^n$. 
The first base vector of the base $B_\phi$ is given as $\frac{\partial}{\partial x_i}|_p := \phi^-1 (\phi(p)+ t e_i)$, the other base vectors of $B_\phi$ follow with $i \in {2,....,n}$.  
Let's call the function that takes $p$ and gives us $\frac{\partial}{\partial x_i}|_p$ and $\frac{\partial}{\partial y_i}|_p$ by names  $\bar{\phi}_p$ and $\bar{\psi}_p$. In a diagram we now got two choices for the base of our ismorphic tangent spaces depending on $\phi,\psi$:
\begin{CD}
   T_{\phi(p)} \mathbb R^n   @. T_{\psi(p)} \mathbb R^n \\
@A \bar{\phi}_p AA @A \bar{\psi}_p AA\\
\mathbb R_\phi^n @. \mathbb R_\psi^n 
\end{CD}
First question: Why is $T_p \mathcal M$ an $\mathbb R$-vector space, i don't know why the underlying field has to be $\mathbb R$. Or is it, that we just identify $T_p M$ with $T_{\phi(p)}\mathbb R^n$, if so, why can we do that?
I have the notion that $\mathcal M$ is of a abstract nature, so how comes it's tangential space is so $\mathbb R^n$ related?
Second question:
Assuming i understood why $T_p \mathcal M$ is a $\mathbb R$-vector space, i can now ask myself how i can change bases.
So we get to linear algebra and the pushforward as linear function between vector spaces.
The image of the coordinate vector $(1,0,...,0)^t$,of the first base vector $\frac{\partial}{\partial x_i}|_p$, is the first row of the matrix $D((\psi)\circ\psi^-1)$ giving us the coordinate vector expressed in the base $B_\psi$.
\begin{CD}
T_{\phi(p)} \mathbb R^n   @>D((\psi)\circ\phi^-1)>> T_{\psi(p)} \mathbb R^n 
\end{CD}
Third question: The "last" notion that gives me a headache right now is that of the function:
$$\phi_*: T_p \mathcal M \to T_{\phi(p)} \mathbb R^n\text{ defined as }
\phi_*(\gamma) := \frac{d}{dt}(\phi \circ \gamma )|_{t=0}$$
where $\gamma$ is a tangent vector, for example a curve(geometric defintion of a tangent vector).
Is that definition base independend? If so taking the first base vector of $B_\phi$
namely $\frac{\partial}{\partial x_i}|_p := \phi^-1 (\phi(p)+ t e_i)$ and using $\phi_*$ on it, gives us another tangent vector in the tangent space $T_{\phi_*} \mathbb R^n$, or is it that same as $\mathbb R_\phi^n$ or $T_{\phi(p)}R^n$?
Thanks for any answers!
 A: First question:
Congratulations, you have discovered algebraic varieties and their Zariski tangent space. For classical differential geometry, the reason that the tangent space is an $\mathbb{R}$ vector space is by definition: the local model of a differentiable manifold is $\mathbb{R}^n$. The whole motivation of classical differential geometry is to study stuff that, at sufficiently small neighborhoods, look like $\mathbb{R}^n$. It would be strange if its tangent space does not look like that of $\mathbb{R}^n$, but be over some other field, no? If you model the local geometry by an affine variety associated to a different base field, you will get appropriately different vector spaces as the tangent space. 
Second question:
Looks like you more or less got it, except you probably meant $\psi\circ \phi^{-1}$ rather than $\psi \circ \psi^{-1}$. Again, locally you are just working with functions defined over $\mathbb{R}^n$ (which may take $\mathbb{R}^k$ values). 
Third question:
$\phi_*$ pushes forward to $T_{\phi(p)}\mathbb{R}^n$, not $T_p\mathbb{R}^n$. Since you didn't explain what $\gamma$ is, I have no idea what your second equation means. But the derivative with respect to $t$ makes me wonder if you intended $\gamma:\mathbb{R}\to M$ and you are actually thinking about the expression
$$ \phi_*(\frac{\mathrm{d}}{\mathrm{d}t}{\gamma}(t)) = \frac{\mathrm{d}}{\mathrm{d}t}( \phi\circ \gamma)(t)~? $$
In that case, this definition is base independent. You can explicitly verify it by composing with the smooth maps associated to changes of charts. 
Your expression $\partial/\partial x_i |_p = \phi^{-1}(\phi(p) + t e_i)$ is wrong. The left hand side is an object living in $T_p M$ and the right hand side is an object living in $M$. You appear to have a type error. 
