algebraic-geometrymultilinear-algebrasheaf-theoryvector-bundles
Given 2 vector bundles $E$ and $F$ of ranks $r_1, r_2$, we can define $k$'th exterior power $\wedge^k (E \otimes F)$.
Is there some simple way to decompose this into tensor products of various exterior powers of individual bundles?
I am interested in the case when $F$ corresponds to the twisted line bundles $\mathcal{O}(k)$.
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.
Yes, that is a fine proof.
As a general principle, if you can write down an isomorphism of vector spaces without making any choices, then you have written down an isomorphism of vector bundles.
Let's make that principle precise in this case. On a small open set $U$, trivialize the vector bundles $V_1$ and $V_2$ with frames (:=sections which form a basis at every point-- note that, seemingly contra the principle, we've made a choice) $e_1, \ldots, e_n$ and $f_1, \ldots, f_m$ (I changed your notation $r_1, r_2$ to $m, n$ to avoid double subscripts).
Then the bundle $V_1 \otimes V_2$ is trivialized with frame $e_i \otimes f_j$ and its determinant trivialized with the singleton frame (alphabetical order) $$ (e_1 \otimes f_1) \wedge (e_1 \otimes f_2) \wedge \ldots \wedge (e_1 \otimes f_m) \wedge(e_2\otimes f_1) \wedge \ldots \wedge (e_n \otimes f_m) $$ which we may map to $$ (e_1\wedge \ldots \wedge e_n)^m \otimes (f_1 \wedge \ldots f_m)^n. $$ Now it seems so far that our map depends on choices, but it does not. We just need to check that multiplying an $e_i$ by an invertible function, adding $\phi e_i$ to $e_j$ (for $\phi$ an arbitrary function), or flip-flopping $e_i$ and $e_j$ does nothing (and same for $f_i$ and $f_j$). Note that in each case, the given bases for $\text{det}(V_1 \otimes V_2)$ and $\text{det} (V_1)^m \otimes \text{det} (V_2)^n$ multiply by the same scalar function, so the map doesn't change.
We were (allegedly) doing all the above reasoning on a small open set $U$ -- otherwise, there may not exist a frame (the existence of a frame on an open set being equivalent to a vector bundle being trivial). Now suppose we define a global map by doing the same reasoning on ALL open sets. We have to check that if $U$ and $W$ are different, we have defined the same map on $U \cap W$.
But it follows from the independence of choices. The restriction of a choice of frame over $U$ to a frame over $U \cap W$ gives the map of bundles over $U \cap W$, and so does the restriction of a choice of frame over $W$. But we know that the map doesn't depend on a choice. So we've therefore given a global map of vector bundles.
In general it can be general to try to work even more abstractly, i.e. not in terms of picking bases, so that it becomes completely automatic, by the above principle, that a map of vector spaces extends to a map of vector bundles.
Best Answer
In the special case when $F$ has rank $1$, the canonical homomorphism $$ (E\otimes F)^{\otimes k}\to (\wedge^k E)\otimes (F^{\otimes k})$$ induces a homomorphism $$ \wedge^k (E\otimes F)\to (\wedge^k E)\otimes (F^{\otimes k}).$$ This is because for all $u\in E$ and all $v_1, v_2\in F$, we have $(u\otimes v_1)\wedge (u\otimes v_2)=0$ (locally $v_i=a_ie$ with $a_i$ scalar and $e$ a basis, so the exterior product vanishes locally, hence vanishes globally). Now again locally (when $F$ is the trivial line the bundle), this homomorphism is clearly an isomorphism. So it is an isomorphism globally: $$ \wedge^k (E\otimes F)\simeq (\wedge^k E)\otimes (F^{\otimes k}).$$