Working out a concrete example of tensor product From this entry in Wikipedia:

The tensor product of two vector spaces $V$ and $W$ over a field $K$ is another vector space over $K$. It is denoted $V\otimes_K W$, or $V\otimes W$ when the underlying field $K$ is understood.
If $V$ has a basis $e_1,\cdots,e_m$ and $W$ has a basis $f_1,\cdots,f_n$, then the tensor product $V\otimes W$ can be taken to be a vector space spanned by a basis consisting of all pairs $(e_i,f_j)$; each such basis element of $V\otimes W$ is denoted $e_i\otimes f_j$.  For any vectors $v=\sum_i v_ie_i\in V$ and $w=\sum_j w_j f_j\in W$ there is a corresponding product vector $v\otimes w$ in $V\otimes W$ given by $\sum_{ij}v_iw_j(e_i\otimes f_j)\in V\otimes W.$  This product operation $\otimes:V\times W \rightarrow V\otimes W$ is quickly verified to be bilinear.
As an example, letting $V=W=\mathbb R^3$ (considered as a vector space over the field of real numbers) and considering the standard basis set $\{\hat x, \hat y,\hat z\}$ for each, the tensor product $V\otimes W$ is spanned by the nine basis vectors 
$\{\hat x \otimes \hat x,\hat x \otimes \hat y,\hat x \otimes \hat z,\hat y \otimes \hat x,\hat y \otimes \hat y, \hat y \otimes \hat z ,\hat z\otimes \hat x,,\hat z \otimes \hat y, \hat z \otimes \hat z  \}$ and is isomorphic to $\mathbb R^9$.  
For vectors $v=(1,2,3),w=(1,0,0)\in \mathbb R^3$ the tensor product $$\bbox[10px, border:2px solid red]{v\otimes w= 
\hat x\otimes \hat x + 2\hat y\otimes \hat x+3\hat z\otimes \hat x}$$
The above definition relies on a choice of basis, which can not be done canonically for a generic vector space.  However, any two choices of basis lead to isomorphic tensor product spaces (c.f. the universal property described below).  Alternatively, the tensor product may be defined in an expressly basis-independent manner as a quotient space of a free vector space over $V\times W$.  This approach is described below.


QUESTION:
If we decide on the standard Euclidean orthonormal basis, what is the final expression of the $v\otimes w$ product in the red boxed expression? Do we eventually get rid of the vector expressions with hats (as well as the $\otimes$ symbols) to get a number as per the (approximate) idea of a tensor as a map from $V\times W\rightarrow \mathbb R?$
What if we change the bases from orthonormal to $\large\begin{bmatrix}\tilde x\\\tilde y\\\tilde z\end{bmatrix}=\begin{bmatrix}3&4&-1\\0&3&7\\1&3&0.5\end{bmatrix}\begin{bmatrix}\hat x\\\hat y\\\hat z\end{bmatrix}?$
 A: Thanks for the hint in comments. It's more clear now what the answer would be:

Applied to the case in the QUESTION, the change of basis matrix is $\small\begin{bmatrix}3&4&-1\\0&3&7\\1&3&0.5\end{bmatrix}$, and its inverse $\small\begin{bmatrix}0.7&0.2&-1.1\\-0.3&-0.1&0.8\\0.1&0.2&-0.3\end{bmatrix}$. The vectors $v$ and $w$ in the new coordinate system are 
$v =\small\begin{bmatrix}0.7&0.2&-1.1\\-0.3&-0.1&0.8\\0.1&0.2&-0.3\end{bmatrix}\begin{bmatrix}1\\2\\3\end{bmatrix} =\begin{bmatrix}-2.3\\1.9\\-0.5\end{bmatrix}$ and $w=\small\begin{bmatrix}0.7&0.2&-1.1\\-0.3&-0.1&0.8\\0.1&0.2&-0.3\end{bmatrix}\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}0.7\\-0.3\\0.1\end{bmatrix}$. 
Therefore,
$$\begin{align}\large v\otimes w=\left(-.23\tilde x + 1.9\tilde y -0.5 \tilde z\right)\otimes \left(0.7\tilde x -0.3\tilde y + 0.1\tilde z\right)\\[2ex]=-1.6\;\tilde x\otimes \tilde x + 1.3\;\tilde x\otimes \tilde y  -0.3 \;\tilde x\otimes \tilde z + 0.6\;\tilde y\otimes \tilde x -0.5\;\tilde y\otimes \tilde y+ 0.1\;\tilde y\otimes \tilde z -0.3\;\tilde z\otimes \tilde x +0.2 \;\tilde z\otimes \tilde y-0.1\;\tilde z\otimes \tilde z\end{align}$$
So what's the point?
Starting off defining the tensor product of two vector spaces ($V\otimes W$) with the same bases, we end up calculating the outer product of two vectors:
$$\large v\otimes_o w=\small \begin{bmatrix}-2.3\\1.9\\-0.5\end{bmatrix}\begin{bmatrix}0.7&-0.3&0.1\end{bmatrix}=\begin{bmatrix}-1.61&0.69&-0.23\\1.33&-0.57&0.19\\-0.35&0.15&-0.05\end{bmatrix}$$
This connect this post to this more general question.
