Consider a ring map $B \rightarrow A$. Consider the map $f:A \otimes_{B}A \rightarrow A$, where $x \otimes y$ goes to $xy$. Let $I$ be the kernel of $f$. Why is it true that $I/I^2$ is isomorphic to $I \otimes_{A \otimes_{B}A} A$?
This is what I've been able to prove till now:
Let $R=A \otimes_{B}A$. Now, consider the $R$-module homomorphism $\phi$from $I$ to $I\otimes (R/I)$, $\phi(a)=a\otimes 1$. It is easy to see that the kernel of $\phi$ contains $I^2$. How do I prove it is exactly $I^2$?
Given an $R$-module $M$ and an ideal $I\subset R$ we have $R$-module morphisms $$M\otimes_R R/I\to M/IM: m\otimes \tilde r\mapsto \overline {rm} \operatorname \quad {and} \quad M/IM \to M\otimes_R R/I:\overline {m}\mapsto m\otimes \tilde 1$$ which are easily seen to be inverse to each other and yield an isomorphism $M\otimes_R R/I\cong M/IM$.
In particular for $M=I$ we get the required isomorphism of $R$-modules $I\otimes_R R/I\cong I/I^2.$
This can be made slightly more general: Let $R$ and $I$ be as you say, and let $M$ be any $R$-module. There is an isomorphism $$ M\otimes_R R/I \cong M/IM. $$ Here's how to see it: start with the exact sequence $$ 0 \to I \to R \overset{f}{\to} R/I \to 0 $$ (note that $f$ is the same as your $f$). Then tensor with $M$: $$ M\otimes_R I \to M\otimes_R R \overset{f\otimes 1}\to M\otimes_R (R/I) \to 0 $$ This is exact by the right-exactness of $\otimes$. Using the natural isomorphism $M \otimes_R R \cong M{}{}{}$, this sequence is exactly $$ M\otimes_R I \overset{\phi}{\to} M \overset{\tilde f}\to M\otimes_R (R/I) \to 0, \qquad (*) $$ where the map $\phi$ takes a pure tensor $m\otimes x$ to $xm$ and $\tilde{f}$ takes $m$ to $m\otimes 1$. It should be clear that the image of $\phi$ is equal to $IM\subset M$.
The nontrivial step that proves the direction you need is the exactness of $(*)$ at the term $M$, which gives $$ IM = \text{image}(\phi) = \ker(M\to M\otimes_R(R/I)). $$ So, in the case that $I=M$, we have that $$ I^2 = \ker(I\to I\otimes_R(R/I)). $$
Unfortunately, I do not know of a good way to prove this by hand. This seems to rely on the right exactness of $\otimes$, and you can see some discussion here that discusses how this is proved.
