I am not sure if this question is better suited for math SE, but given that I am asking for a geometrical (ideally physical) interpretation, I figured it would be best asked here.
I am looking for a way to understand tensor product spaces intuitively. I understand both the basis-dependent and the algebraic basis-free definitions. What I am struggling with is interpreting what something of the form $$v \otimes w, \quad v \in V, w \in W$$ means geometrically, where $V$ and $W$ are vector spaces.
For example, in $\mathbb{R}^3$ it is very easy to explain and visualize what any vector or basis looks like. For an arbitrary $x \in \mathbb{R}^3$ we can decompose it into $$x = a_1 e_1 + a_2 e_2 + a_3e_3$$ and this geometrically makes clear sense, we are simply breaking down the vector into its component parts. In other words, if $x$ denotes the position of some particle we can say it is $a_i$ distance in the $i$th direction ($e_i)$.
However, in the case of tensor products, does a similar geometric interpretation hold? How does one exactly interpret an arbitrary vector $v \otimes w$? To make the question simpler, take $V = \mathbb{R}^2$ (with basis $\{e_i^1\}$) and $W = \mathbb{R}^3$ (with basis $\{e_i^2\})$. Then $$\begin{align}v \otimes w = &a_1 e^1_1\otimes e_1^2 + a_2 e^1_2\otimes e_1^2 + a_3 e^1_1\otimes e_2^2 + \\&a_4 e^1_2\otimes e_2^2 + a_5 e^1_1\otimes e_3^2 + a_6 e^1_2\otimes e_3^2 \end{align}$$
The only picture I am able to have in mind is in the context of quantum mechanics. If the two vector spaces represent different observable quantities, then I interpret the tensor product as taking into account two things at once. But this is unsatisfying to me.
