If $A$, $B$ and $C$ are $k$-algebras (where $k$ is a commutative ring), then the general rule for tensor products if the following: for an $A$-$B$-bimodule $_A M_B$ and a $B$-$C$-bimodule $_BN_C$, the tensor product $M\otimes_B N$ is an $A$-$C$-bimodule.
Let's apply this to your situation. You have a finite-dimensional $k$-algebra, say $\Lambda$. You have two modules over $\Lambda$. Let's say they are left modules $_\Lambda M$ and $_\Lambda N$. Since none of them is a right $\Lambda$-module, we cannot take their tensor product over $\Lambda$. The only reasonable choice is to take their tensor product over $k$.
To do this, note that $M$ has a natural structure of right $k$-module, and that $N$ has a structure of left $k$-module. Thus $M\otimes_k N$ is defined. According to the above, it is a left $\Lambda$-module; however, it is not so interesting, since it is isomorphic to a direct sum of a number of copies of $M$ equal to the $k$-dimension of $N$.