What is it and how 'huge' is really the formal product of vector spaces? In here the formal product of two vector spaces $V$ and $W$ is defined as the span of symbols $v \ast w$
$$V \ast W = \text{span}\quad\{v \ast w \mid v,w \in \mathbb R \}$$
I guess the wording "span of symbols" implies a completely abstract definition and excludes the possibility of finding a basis, and hence, the statement that the dimension of this space is 'huge'.
Does that imply that each object so constructed would be its implicit basis, such that
$$\begin{pmatrix}1\\2\end{pmatrix} \ast \begin{pmatrix}-1\\2\\\pi\end{pmatrix}$$
would contribute $1$ to the dimensionality, as much as the object
$$\begin{pmatrix}-1\\ -2\end{pmatrix} \ast \begin{pmatrix}-1\\2\\\pi\end{pmatrix}$$
would also increase the dimensionality by $1$?
If so would the cardinality of the set be larger than the reals?
 A: 
excludes the possibility of finding a basis

Every vector space has a basis (assuming the axiom of choice) and this one is no exception. But the basis consists of all pairs of vectors from the original vector spaces. As you say, each pair of vectors "contributes one dimension" so for the case in the video of $V= \mathbb{R}^2$ and $W=\mathbb{R}^3$, $V *W$'s dimension is uncountably infinite (the cardinality of $\mathbb{R}^5$).
In particular, it's not true that
$$\left(\begin{bmatrix} 2 \\ 2\end{bmatrix}*\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}\right) = 2\left(\begin{bmatrix} 1 \\ 1\end{bmatrix}*\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}\right)$$
nor is it true that
$$\left(\begin{bmatrix} 1 \\ 1\end{bmatrix}*\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}\right) = \left(\begin{bmatrix} 0 \\ 1\end{bmatrix}*\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}\right)+ \left(\begin{bmatrix} 1 \\ 0\end{bmatrix}*\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}\right).$$
The "point" of the tensor product is that it quotients out $V * W$ by the symmetries needed to make $V\otimes W$ a vector space spanned by pairs of basis vectors of $V$ and $W$ only.
