By the universal properties of quotient algebras, tensor products ( = coproducts of algebras), we have for every $R$-algebra $T$:
$\hom(A/I \otimes B/J,T) \cong \hom(A/I,T) \times \hom(B/J,T)$
$\cong \{f \in \hom(A,T),g \in \hom(B,T) : f|_I = 0, g|_J = 0\}$
$\cong \{h \in \hom(A \otimes_R B,T) : f:=h(- \otimes 1), g:=h(1 \otimes -) \text{ satisfy } f|_I = 0,~ g|_J = 0\}$
$\cong \{h \in \hom(A \otimes_R B,T) : h|_{I \otimes 1 + 1 \otimes J}=0\}$
$\cong \hom((A \otimes_R B)/(I \otimes 1 + 1 \otimes J),T)$
By the Yoneda lemma, we are done.
Remark: This is one of the thousands of trivial isomorphisms in basic algebra which are usually proved (in textbooks, lectures, etc.) in a too complicated way. Instead, you can always just use the Yoneda lemma and the involved universal properties. And then there is nothing to do at all ... By the way, this abstract approach is the only one which is applicable in more abstract contexts, where you can't use elements anyway.
For an inverse, define $M \to R \otimes M$ by $m \mapsto 1 \otimes m$. I suppose you could also show directly that the map is injective, but point remains the same: $R$ contains $1$. If $\sum r_i \otimes m_i$ is in the kernel then $\sum r_im_i = 0$, and we can write the tensor as $\sum 1 \otimes r_im_i = 1 \otimes \sum r_im_i = 1 \otimes 0 = 0$.
Another way is to show that your map $R \times M \to M$ satisfies the universal property of the tensor product of $R$ and $M$.
Best Answer
Consider the map $g: R/(I+J)\to R/I \otimes_R R/J$ given by $\overline{a} \mapsto \overline{a} \otimes 1$. If you confirm for yourself that this is a well-defined function, it's obvious that it's an inverse to your map $f$.