Why does $(A/I)\otimes_R (B/J)\cong(A\otimes_R B)/(I\otimes_R 1+1\otimes_R J)$?

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.

Let's solve your problem in two steps:

Step 1
Consider the exact sequence of $R$-modules $0\to I\to A\to A/I \to 0$ .
Tensoring with $B$ and remembering that tensoring is right exact we obtain
the exact sequence $I\otimes_R B \to A\otimes_R B\to A/I\otimes_R B\to 0$.
Writing $I^e$ for the image of $I\otimes_R B \to A\otimes_R B \;$ [the exponent e in $I^e$ stands for "extension of ideal $I$ to ring $A\otimes_R B $" ] we get the identification of $R$-algebras $$A/I\otimes_R B\cong (A \otimes_R B)/I^e:\bar a\otimes b \mapsto \overline {a \otimes b} \quad (*)$$

Step 2
Applying the corresponding result in Step 1 to the right hand side of the tensor product we get $$A/I\otimes_R B/J\cong (A/I\otimes_R B)/J^e \quad (**)$$
Applying Step 1 again, we replace $A/I\otimes_R B$ in $(**)$ by $(A\otimes_R B)/I^e$ and get
$$(A/I\otimes_R B)/J^e \cong \frac {(A\otimes_R B)/I^e}{I^e+J^e/I^e} \quad (***)$$ where the ideal $I^e +J^e$ , denoted $I\otimes 1+1\otimes J$ in Danielle's question, is the subgroup of $A\otimes_R B$ generated by elements of the form $i\otimes b+a\otimes j$ where $i\in I, b\in B, j\in J,\, a\in A$

Conclusion
Finally thanks to Noether we obtain from $(**)$ and $(***) $ the required final identification of $R$-algebras : $$ A/I\otimes_R B/J\cong (A\otimes_R B)/(I^e +J^e):\bar a\otimes \bar b\mapsto \overline {a\otimes b} \quad (****) $$

Think about $$ A\times B \rightarrow A/I \otimes B/J \quad\text{induced from}\quad A \times B \rightarrow A/J \quad \text{and} \quad A\times B \rightarrow B/J $$ giving you a homomorphism $$ \Phi:A \otimes B \rightarrow A/I \otimes B/J \quad \Phi(a\otimes b)=(a+I)\otimes(b+J) $$ It is obvious that $I\otimes 1 \subseteq \operatorname{Ker}\Phi$ and $1\otimes J \subseteq \operatorname{Ker}\Phi$ so their sum is also in the kernel. It remains to show that the sum is the kernel.

Note that $I\otimes 1$ is the ideal generated by $f(I)$ for $f:A\rightarrow A\otimes B$ and $1\otimes J$ is the ideal generated by $g(J)$ for $g:B \rightarrow A\otimes B$. Then there are homomorphisms $$ A/I \rightarrow A\otimes B/(f(I)+g(J)) \quad \text{and} \quad B/J \rightarrow A\otimes B/(f(I)+g(J)) $$ and hence $$ A/I \otimes B/J \rightarrow A\otimes B/(f(I)+g(J)) \rightarrow A\otimes B/\operatorname{Ker}(\Phi) \rightarrow A/I \otimes B/J $$ Can you show that $f^{-1}(f(I)+g(J))=I$ and $g^{-1}(f(I)+g(J))=J$?