I am proving that $\mathrm{Tor}_1^R(R/I,R/J)=(I\cap J)/(IJ)$. For that, I consider the exact sequence $0\rightarrow I \rightarrow R\rightarrow R/I\rightarrow 0$, and then I have an long exact sequence
$\begin{align}\DeclareMathOperator{\Tor}{Tor} \dots\rightarrow0=\Tor^R_1(R,R/J)\rightarrow\Tor^R_1(R/I,R/J)\rightarrow I\otimes_R R/J \rightarrow R/J\rightarrow R/I\otimes_R R/J \rightarrow 0 \end{align}$
By that, to conclude I only need to prove that the map $ \Tor^R_1(R/I,R/J)\rightarrow I\otimes_R R/J $ is surjective, because in that case by the exactness of the sequence I have the result.
¿Is that map necessary surjective?
By definition of exact sequence, $\operatorname{Tor}^R_1(R/I,R/J)$ is the kernel of the map $I\otimes_RR/J\to R/J$.
On the other hand, you also have the exact sequence $0\to J\to R\to R/J\to0$ and tensoring with $I$ you get $$ I\otimes_RJ\to I\otimes_RR\to I\otimes_R R/J\to 0 $$ so $I\otimes_R R/J\cong I/K$ and it's not difficult to show that $K=IJ$. Then check that the map $I\otimes_RR/J\to R/J$ translates to the map $$ I/IJ\to R/J, \quad a+IJ\mapsto a+J $$ whose kernel is $(I\cap J)/IJ$.