Consider the short exact sequence $0 \to I \to A \to A/I \to 0$. Tensoring with the $A$-module $A/J$ gives the long exact sequence $\cdots \to 0 \to \mathrm{Tor}_2^A(A/I,A/J) \to \mathrm{Tor}_1^A(I,A/J) \to 0 \to \mathrm{Tor}_1^A(A/I,A/J) \to I/IJ \to A/J \to (A/I)\otimes_A (A/J)\to 0$. Here I use the fact that $A$ is a free $A$ module, so all its Tor's are 0. The right part of the sequence gives the first equality you mention. The left part identifies $\mathrm{Tor}_2^A(A/I,A/J)$ with $\mathrm{Tor}_1^A(I,A/J)$.

Now tensor $0 \to J \to A \to A/J \to 0$ with the $A$-module $I$. This gives $\cdots \to 0 \to \mathrm{Tor}_1^A(A/J,I) \to J\otimes_A I \to I \to I/IJ \to 0$, so $0 \to \mathrm{Tor}_1^A(A/J,I) \to J\otimes_A I \to IJ \to 0$, and since Tor and $\otimes$ are symmetric, $\mathrm{Tor}_2^A(A/I,A/J)$ is the kernel of $I\otimes_A J \to IJ$.