Tensor Product of Spaces has Basis of Tensor Products I am given the following definition of the Tensor Product of spaces
Given two vector spaces $V,W$ a vector space S is a tensor product of $V,W$ if there exists a map $M$ 
$$ M: V \times W \rightarrow S$$ 
Such that $M(x_1,x_2)$ is linear in $x_1$ and $x_2$ if $(e_1)$ is a basis of $V$ and $(e_2)$ a basis of $W$ then $M(e_1,e_2)$ is a basis of S.
Then we say that $M = V \otimes W$
So now this is certainly a definition I understand and can work with but I want to build a picture in my head about what is going on. 
I know of the traditional tensor product that is given two matrices
$$A \otimes B= \begin{bmatrix} a_{00}B \ \ a_{01}B \ \ ... \ \ a_{0m}B \\  a_{10}B \ \ a_{11}B \ \ ... \ \ a_{1m}B \\ \vdots \ \ \vdots  \ \ \ddots  \ \ \vdots \\ a_{n0}B \ \ a_{n1}B \ \ ... \ \ a_{nm}B \end{bmatrix} $$
So is it safe to say that given a basis $(e_1)$ of $V$ and and $(e_2)$ of W that all the possible tensor products $uv, u \in (e_1), v \in (e_2)$ is the basis of S?
Looking at: https://en.wikipedia.org/wiki/Tensor_product#Tensor_product_of_vector_spaces
didn't yield too much insight, as they seemed to be covering a much more general and abstract formulation that (might answer my question but) felt like overkill for my needs.
 A: Note that usually we'll write $x_1 \otimes x_2$ instead of $M(x_1,x_2)$. (Your $M$ is the canonical map $M:V \times W \to V \otimes W$ that sends a pair to the underlying tensor.)
Matrices correspond to linear maps, and the tensor product of matrices corresponds to the tensor product of linear maps. Given two linear maps $f_i:V_i \to W_i$, their tensor product $f_1 \otimes f_2:V_1 \otimes V_2 \to W_1 \otimes W_2$ is defined by $(f_1\otimes f_2)(v_1,v_2) := f_1(v_1) \otimes f_2(v_2).$ Then what holds is that the matrix corresponding to $f_1 \otimes f_2$ (with respect to the tensor product of the chosen bases) is precisely the tensor product of the matrices corresponding to $f_1$ and $f_2$.
Example: let $f_1:\mathbb{R} \to \mathbb{R}^2$ be given by $f(a) = (a,0)$ and let $f_2:\mathbb{R}^2 \to \mathbb{R}$ be given by $f_2(a,b) = a$. Choose the standard bases for $\mathbb{R}$ and $\mathbb{R}^2$. The matrix of $f_1$ is $A = \begin{pmatrix}1 \\0 \end{pmatrix}$ and the matrix of $f_2$ is $B = \begin{pmatrix}1 &0 \end{pmatrix}$. We have $f_1 \otimes f_2:\mathbb{R} \otimes \mathbb{R}^2 \to \mathbb{R}^2 \otimes \mathbb{R}$ defined by $(f_1 \otimes f_2)(a \otimes (b,c)) = f_1(a) \otimes f_2(b,c) = (a,0) \otimes b$. Thus in the bases $(1 \otimes (1,0), 1 \otimes (0,1))$ for $\mathbb{R} \otimes \mathbb{R}^2$ and $((1,0) \otimes 1, (0,1)\otimes 1)$ for $\mathbb{R}^2 \otimes \mathbb{R}$, we see that the matrix of $f_1 \otimes f_2$ really is $A \otimes B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$.
