How to prove that $R/I \otimes_R M \cong M / IM$

Consider $$ 0 \to I \to R \to R/I \to 0 $$ Apply the right exact functor $-\otimes_R M$, and you get $$ I\otimes_R M \to R\otimes_R M \to (R/I)\otimes_R M \to 0 $$ But $R\otimes_R M$ is canonically identified with $M$ by $a\otimes m \mapsto am$. Then $I\otimes_R M \to R\otimes_R M = M$ is $a\otimes m \mapsto am$ so, by definition, its image is $IM$. So the exactness of the sequence tells you that
$$ M/IM \cong (R/I)\otimes_R M $$

In general, playing around with elements of a tensor product is something that should be avoided if you can help it (defining maps from a tensor product is also usually better done by defining them from the product and then checking that they are in fact $R$-balanced and the using the universal property). A proof which avoids dealing with explicit elements of the tensor product can be done in the following way:

Consider the map $R/I\times M\to M/IM$ given by $(r+I,m)\mapsto rm+IM$. This is well-defined, as the image of any two representatives of $r+I$ differ by an element of $IM$. Since it is $R$-balanced, it descends to a map from the tensor product. Now consider the map $M \to R/I\otimes M$ given by sending $m\mapsto (1+I)\otimes m$. This descends to the quotient $M/IM$ by the 1st isomorphism theorem, and then it is easy to see that these maps are mutual inverses.