How do you "restrict" a smooth tensor field to a point? I am confused about some aspects of the following definition.

Let $M$ be a smooth manifold, $C^{\infty}\left(M\right)$ denote the commutative ring of smoth functions on $M$ and $C^{\infty}\left(TM\right)$ be the set of smoth vector fields on $M$ forming a module over $C^{\infty}\left(M\right).$ 
  Put $C_0^{\infty}\left(TM\right)=C^{\infty}\left(M\right)$ and for each positive integer $r$ let
  $$C_r^{\infty}\left(TM\right)= C^{\infty}\left(TM\right) \otimes \ldots \otimes C^{\infty}\left(TM\right)$$
  be $r$-fold tensor product of $C^{\infty}\left(TM\right)$ over $C^{\infty}\left(M\right).$
Definition:
  Let $M$ be a differentiable manifold. A smooth tensor field $T$ on $M$ of type $\left(r,s\right)$ is a map $T:C_r^{\infty}\left(TM\right) \rightarrow C_s^{\infty}\left(TM\right)$ which is multi-linear over $C^{\infty}\left(M\right)$ i.e.
  $$T\left( X_1 \otimes \dots X_{k-1} \otimes \left( f.Y+g.Z \right) \otimes X_{k+1} \otimes \dots \otimes X_r \right) $$
  $$=f.T\left( X_1 \otimes \dots X_{k-1} \otimes Y  \otimes X_{k+1} \otimes \dots \otimes X_r \right)+g.T\left( X_1 \otimes \dots X_{k-1} \otimes Z  \otimes X_{k+1} \otimes \dots \otimes X_r \right)$$
  for all $X_1, \dots, X_r,Y,Z \in C^{\infty}\left(TM\right)$ and $f,g \in C^{\infty}\left(M\right).$
Alternatively, we use the notation $T\left( X_1 , \dots , X_r \right)$ for $T\left( X_1 \otimes \dots \otimes X_r \right).$
We denote the multi-linear restriction of $T$ to the $r-$fold tensor product $TM_p \otimes \dots \otimes TM_p$ of the vector space $TM_p$ over $\mathbb{R}$ by $T_p : \left( X_1\left( p \right) , \dots , X_r\left( p \right) \right)  \mapsto T\left( X_1 , \dots , X_r \right) \left(p\right). $

I cannot understand that why $T$ can be restricted to the $r-$fold tensor product $TM_p \otimes \dots \otimes TM_p$ of the vector space $TM_p$ over $\mathbb{R}.$ I am wondering what the domain of $T_p$ is? If its domain is $TM_p \otimes \dots \otimes TM_p$ then $C_r^{\infty}\left(TM\right)$ has to contain $TM_p \otimes \dots \otimes TM_p$? I do not know whether or not this is true?
On the other hand, if its domain is $C_r^{\infty}\left(T_pM\right)$ (we consider on tangent space not tangent bundle) then $C_r^{\infty}\left(T_pM\right)$ is a set of vector fields which go through $p$? However, it is necessary to define the Riemannian metric under $T_p$ which has domain is $T_pM \otimes T_pM.$
I should be very glad to recieve responses from anyone reading this who have comments or suggestions of decent!
 A: It is indeed inaccurate to call $T_p$ the "restriction" of $T$, since $TM_p\otimes\dots\otimes TM_p$ is not a subset of $C^\infty_r(TM)$ in any natural way.  Instead, it is a quotient: given an element of $TM$, you can get an element of $TM_p$ by evaluating at $p$.  Given an element of $C^\infty_r(TM)$, you can then get an element of $TM_p\otimes\dots\otimes TM_p$ by evaluating each of the vector fields in your tensor at $p$.
So if it doesn't make sense to restrict $T$, how is $T_p$ defined?  The answer is as follows.  Suppose $v_1,\dots,v_r\in TM_p$.  Then we can choose vector fields $X_1,\dots,X_r\in TM$ such that $X_i(p)=v_i$ for $i=1,\dots,r$.  We then define $$T_p(v_1\otimes\dots\otimes v_r)=T(X_1\otimes\dots\otimes X_r)(p).$$  (And if you check that this is multilinear in $v_1,\dots,v_r$, this uniquely extends to a map $T_p$ defined on $TM_p\otimes\dots\otimes TM_p$.)
Now, there's an obvious problem with this definition: there are lots of different choices of $X_1,\dots,X_r$ you could make, and it's not obvious that the definition of $T_p(v_1\otimes\dots\otimes v_r)$ ends up being the same no matter what choices you make.  However, it turns out that the choice of $X_1,\dots X_r$ doesn't make a difference, because of the assumption that $T$ is $C^\infty(M)$-linear.  For simplicity, let me illustrate how this works in the case $M=\mathbb{R}$ and $r=1$; the general case is similar but more complicated.
So we have a tangent vector $v\in TM_p$, and want to prove that if $X$ and $Y$ are two vector fields such that $X(p)=Y(p)=v$, then $T(X)(p)=T(Y)(p)$ (so that we can define $T_p(v)=T(X)(p)$ and this won't depend on the choice of $X$).  Let $Z=X-Y$, so $Z(p)=0$; it suffices to show that $T(Z)(p)=0$.  Now since we are assuming $M=\mathbb{R}$, we can think of this vector field $Z$ as just a smooth function $\mathbb{R}\to\mathbb{R}$.  Now define $W(t)=Z(t)/(t-p)$ for $t\neq 0$ and $W(p)=Z'(p)$.  Since $Z$ is smooth and $Z(p)=0$, $W$ is also smooth, and we have $Z=f\cdot W$, where $f(t)=t-p$.
Now we use the $C^\infty(M)$-linearity of $T$.  Since $Z=f\cdot W$, $T(Z)=f\cdot T(W).$  Evaluating at $p$, we get $T(Z)(p)=f(p)T(W)(p)$.  But $f(p)=p-p=0$, so we conclude that $T(Z)(p)=0$, as desired.
(When $M=\mathbb{R}^n$ rather than just $\mathbb{R}$, you need to use Hadamard's lemma to get an expression analogous to this expression $Z=f\cdot W$ in order to show that $T(Z)(p)=0$: if $p=(p_1,\dots,p_n)\in\mathbb{R}^n$, then it is possible to find $W_1,\dots,W_n$ such that $$Z(t_1,\dots,t_n)=(t_1-p_1)W_1(t_1,\dots,t_n)+\dots+(t_n-p_n)W_n(t_1,\dots,t_n).$$  In the case that $M$ is a general manifold, you need to use a bump function on a coordinate chart near $p$ to reduce to the case of $\mathbb{R}^n$.)
