As title indicates. This is actually a partial exercise in MacDonalds. But most of these relevant exercises boils down to constructing some natural map and using universal property to argue the isomorphisms.
Anyways my problem is this very loose statement that $M \otimes_A N$ is actually a $B-$module.
So I thought I just write out the free module that would be generated by $M \times N$, $C = \{ \sum_{i,j = 1}^n a_ib_j(m_i,n_{ij}) : n = |M \times N|\}$. Now I feel like the answering is staring at me in this sum, but I can't draw out the answer atm. I honestly think the answer lies in the part "$b_j(m_i$" where the $b_js$ are "touching" the $m_i$ is the reason.