Let $A, B$ be $k$-algebras. $M$ is $A$-$B$-bimodule that is finitely generated projective as a left $A$-module and as a right $B$-module, $N, U$ are $B$-$A$-bimodules that is finitely generated projective as a left $B$-module and as a right $A$-module.
As a left $B$-module, $U$ is a direct summand of $B^n$ for some positive integer $n$, hence $M ⊗_B U$ is a direct summand of $M^n$ as a left $A$-module. This shows that $M ⊗_B U$ is fnitely generated projective as a left $A$-module. A similar argument shows that $M ⊗_B U$ is finitely generated projective as a right $A$-module.
I can't see how a similar argument works. I tried and found I cannot tensor with M anymore as $A$ is not a $B$-module. Thank you!
You want to show that $M\otimes_B U$ is finitely generated projective as a right $A$-module. I suppose you can see it is finitely generated, so let us check that the functor $$X\longmapsto\hom_A(M\otimes_B U,X)$$ is exact. By adjunction, this functor is equal to $$X\longmapsto\hom_B(M,\hom_A(U,X))$$ and this is a composition of two functors that are exact, since $M$ and $U$ are projective.