This is a generalization of the currently unanswered question here.
Let $k$ be a field, $A$ be a finite-dimensional $k$-algebra, and $M$, $N$ right and left $A$-modules, respectively, both finite dimensional over $k$. How can one compute the dimension (over $k$) of the tensor product $M\otimes_A N$?
Let's say $A$ has dimension $a$, $M$ has dimension $m$ and $N$ has dimension $n$. It is not too difficult to see that $M\otimes_A N$ is a quotient of $M\otimes_k N$, therefore we have $\dim_k(M\otimes_A N)\leq mn$. Moreover, there is a short exact sequence $$0\to Z\to M\otimes_k N\to M\otimes_A N \to 0$$ so identifying $Z$ and its dimension would be sufficient: $\dim_k(M\otimes_A N) = mn - \dim_k(Z)$. Giving an explicit description of $Z$ would be a satisfying answer for me.
One can easily find counterexamples that show the answer does not simply depend on $m$, $n$, and $a$: Let $k = \mathbb{C}$, $A = \mathbb{C}[x]/(x^2)$, $M = \mathbb{C}^2$ and $N = \mathbb{C}^2$ (where $x$ acts by 0). Then we have $\dim_k(A\otimes_A A) = \dim_k(A) = 2$ but $\dim_k(M\otimes_A N) = 4$ even though $M, N$ and $A$ are all $2$-dimensional. Therefore we are forced to look at the specific actions of $A$ on $M$ and $N$.
Perhaps one can extract something from the implied maps $A\to \text{End}_k(M)$ and $A\to \text{End}_k(N)$.