Solution 1:

Let $f \in \operatorname{Hom}_R(M\otimes_S N,Q)$. We define $g \in \operatorname{Hom}_S(M, \operatorname{Hom}_R(N,Q))$ by:

$$g(m)(n)=f(m \otimes n)$$

Similarly, if $g$ is defined, we can easily define $f$.

I'll leave it to you to prove that this map between $f$ and $g$ actually goes to the appropriate sets, but this is the basic argument. As mentioned by one of the comments, it does depend on how you define the tensor product.