Main question:
Is there any abstract explanation of why the tensor (Kronecker) product of maps turns up in the situation given below?
I was working on the same question from Halmos as this one, which is:
Let $A_1, \ldots, A_n:V\to V$ be linear transformations of an $n$-dimensional vector space $V$. Let $T=(\alpha_{ij})_{ij}$ be an invertible $n$ by $n$ matrix. Then all the $A_i$ commute if and only if all the $\sum_k \alpha_{ik}A_k$ commute.
(The interesting direction is of course the "if"). My solution was more nitty-gritty than the answer in the link, and I admit that I don't completely follow their last inference... What I did was to consider all the equations $$\left(\sum_k \alpha_{ik}A_k\right)\left(\sum_k \alpha_{jk}A_k\right) = \left(\sum_k \alpha_{jk}A_k\right)\left(\sum_k \alpha_{ik}A_k\right).$$ Expanding and simplifying, it turned out by inspection of the coefficients that this could be written as $$ (T\otimes T)([A_i,A_j])_{ij} = 0, $$ where $T\otimes T$ is the Kronecker (or tensor) product, and $([A_i,A_j])_{ij}=(A_iA_j-A_jA_i)_{ij}$ is a vector of commutators running over all ordered pairs $1\le i,j\le n$.* Now $T\otimes T$ is invertible since $T$ is invertible, hence $([A_i,A_j])_{ij} = 0$, and we are done.
So, is the above equation related to a tensor product at all? The abstract tensor product of transformations on $V$ should be a transformation on $V\otimes V$, i.e. it should have tensors as input and output. Perhaps Is it perhaps the case that if $A$ and $B$ are vectors in the space of transformations on $V$, then the tensor $A\otimes B$ can be identified with $AB$? I at least note that $(A,B)\mapsto AB$ is bilinear, so it seems plausible enough to me at first glance.
*Well, I guess a vector needs scalars strictly speaking, so you can think of $([A_i,A_j])$ as a matrix of row vectors in some basis in the space of linear transformation on $V$.
As a small aside, we can also compress the system of equations found, since we always have $[A_i,A_i]=0$ and $[A_i,A_j]=-[A_j,A_i]$. Let $\beta_{ij,kl} = \alpha_{ik}\alpha_{jl}-\alpha_{jk}\alpha_{il}$. Then (barring typos), we should have $$ (\beta_{ij,kl})_{ij,kl}([A_i,A_j])_{ij}, $$ where we now index over unordered distinct pairs $1\le i<j\le n$ and $1\le k<l\le n$. This seemed more difficult to work with for me, but I included it here in case it could lead to some further insight. I suspect that $(\beta_{ij,kl})_{ij,kl}$ is probably an invertible matrix as well. Note that $\beta_{ij,kl}$ is a two by two minor of the matrix $T$, which I guess could be significant.
Consider the linear map
$$F : K^n \ni (x_1, \dots x_n) \mapsto \sum_{i=1}^n x_i A_i \in \text{End}(V).$$
The commutator is a "bilinear form" $[-, -] : \text{End}(V) \times \text{End}(V) \to \text{End}(V)$, and by bilinearity it follows that the $A_i$ commute iff the commutator is identically zero on the image of $F$, or equivalently iff
$$[F(x), F(y)] = 0$$
for all $x, y \in K^n$. Phrasing it this way makes it clear that this condition does not depend on a choice of basis of $K^n$ (it depends only on the image of $F$), so we can switch to another one, or equivalently we can precompose $F$ with any automorphism of $K^n$; this is our invertible matrix $T$, and this is an abstract way to solve the exercise.
The relevance of the tensor product comes from the bilinearity of the commutator. Abstractly $[F(x), F(y)]$ can be constructed by first taking the tensor square of $F$, producing $F^{\otimes 2} : (K^n)^{\otimes 2} \to \text{End}(V)^{\otimes 2}$, then composing with the commutator thought of as a map $\text{End}(V)^{\otimes 2} \to \text{End}(V)$. The need to take the tensor square of $F$ here is what is responsible for the appearance of $T^{\otimes 2}$ when we precompose with $T$; more explicitly, $[F(T(x)), F(T(y))]$ is the composite
$$(K^n)^{\otimes 2} \xrightarrow{T^{\otimes 2}} (K^n)^{\otimes 2} \xrightarrow{F^{\otimes 2}} \text{End}(V)^{\otimes 2} \xrightarrow{[-, -]} \text{End}(V).$$
(Possibly we need to replace $T$ with some combination of its inverse or its transpose to make everything match up but it doesn't really affect the abstract content of the argument.)