First of all, I am very comfortable with the tensor product of vector spaces. I am also very familiar with the well-known generalizations, in particular the theory of monoidal categories. I have gained quite some intuition for tensor products and can work with them. Therefore, my question is not about the definition of tensor products, nor is it about its properties. It is rather about the mental images. My intuition for tensor products was never really geometric. Well, except for the tensor product of commutative algebras, which corresponds to the fiber product of the corresponding affine schemes. But let's just stick to real vector spaces here, for which I have some geometric intuition, for example from classical analytic geometry.
The direct product of two (or more) vector spaces is quite easy to imagine: There are two (or more) "directions" or "dimensions" in which we "insert" the vectors of the individual vector spaces. For example, the direct product of a line with a plane is a three-dimensional space.
The exterior algebra of a vector space consists of "blades", as is nicely explained in the Wikipedia article.
Now what about the tensor product of two finite-dimensional real vector spaces $V,W$? Of course $V \otimes W$ is a direct product of $\dim(V)$ copies of $W$, but this description is not intrinsic, and also it doesn't really incorporate the symmetry $V \otimes W \cong W \otimes V$. How can we describe $V \otimes W$ geometrically in terms of $V$ and $W$? This description should be intrinsic and symmetric.
Note that SE/115630 basically asked the same, but received no actual answer. The answer given at SE/309838 discusses where tensor products are used in differential geometry for more abstract notions such as tensor fields and tensor bundles, but this doesn't answer the question either. (Even if my question gets closed as a duplicate, then I hope that the other questions receive more attention and answers.)
More generally, I would like to ask for a geometric picture of the tensor product of two vector bundles on nice topological spaces. For example, tensoring with a line bundle is some kind of twisting. But this is still some kind of vague. For example, consider the Möbius strip on the circle $S^1$, and pull it back to the torus $S^1 \times S^1$ along the first projection. Do the same with the second projection, and then tensor both. We get a line bundle on the torus, okay, but how does it look like geometrically?
Perhaps the following related question is easier to answer: Assume we have a geometric understanding of two linear maps $f : \mathbb{R}^n \to \mathbb{R}^m$, $g : \mathbb{R}^{n'} \to \mathbb{R}^{m'}$. Then, how can we imagine their tensor product $f \otimes g : \mathbb{R}^n \otimes \mathbb{R}^{n'} \to \mathbb{R}^m \otimes \mathbb{R}^{m'}$ or the corresponding linear map $\mathbb{R}^{n n'} \to \mathbb{R}^{m m'}$ geometrically? This is connected to the question about vector bundles via their cocycle description.
Best Answer
Well, this may not qualify as "geometric intuition for the tensor product", but I can offer some insight into the tensor product of line bundles.
A line bundle is a very simple thing -- all that you can "do" with a line is flip it over, which means that in some basic sense, the Möbius strip is the only really nontrivial line bundle. If you want to understand a line bundle, all you need to understand is where the Möbius strips are.
More precisely, if $X$ is a line bundle over a base space $B$, and $C$ is a closed curve in $B$, then the preimage of $C$ in $X$ is a line bundle over a circle, and is therefore either a cylinder or a Möbius strip. Thus, a line bundle defines a function $$ \varphi\colon \;\pi_1(B)\; \to \;\{-1,+1\} $$ where $\varphi$ maps a loop to $-1$ if its preimage is a Möbius strip, and maps a loop to $+1$ if its preimage is a cylinder.
It's not too hard to see that $\varphi$ is actually a homomorphism, where $\{-1,+1\}$ forms a group under multiplication. This homomorphism completely determines the line bundle, and there are no restrictions on the function $\varphi$ beyond the fact that it must be a homomorphism. This makes it easy to classify line bundles on a given space.
Now, if $\varphi$ and $\psi$ are the homomorphisms corresponding to two line bundles, then the tensor product of the bundles corresponds to the algebraic product of $\varphi$ and $\psi$, i.e. the homomorphism $\varphi\psi$ defined by $$ (\varphi\psi)(\alpha) \;=\; \varphi(\alpha)\,\psi(\alpha). $$ Thus, the tensor product of two bundles only "flips" the line along the curve $C$ if exactly one of $\varphi$ and $\psi$ flip the line (since $-1\times+1 = -1$).
In the example you give involving the torus, one of the pullbacks flips the line as you go around in the longitudinal direction, and the other flips the line as you around in the meridional direction:
Therefore, the tensor product will flip the line when you go around in either direction:
So this gives a geometric picture of the tensor product in this case.
Incidentally, it turns out that the following things are all really the same:
Line bundles over a space $B$
Homomorphisms from $\pi_1(X)$ to $\mathbb{Z}/2$.
Elements of $H^1(B,\mathbb{Z}/2)$.
In particular, every line bundle corresponds to an element of $H^1(B,\mathbb{Z}/2)$. This is called the Stiefel-Whitney class for the line bundle, and is a simple example of a characteristic class.
Edit: As Martin Brandenburg points out, the above classification of line bundles does not work for arbitrary spaces $B$, but does work in the case where $B$ is a CW complex.