If $S \supset R$ is an integral extension of commutative rings, $1_S \in R$, and $I \subset S$ an ideal in $S$, then I can see that since $\exists f \in R[x]$ such that $f(s) = 0$, $\forall s \in S$, then $\bar{f}(\bar{s}) = 0$ where $\bar{f} \in R/(R\cap I)[x]$, but I'm not quite seeing how $R/(R\cap I) \subset S/I$ as rings.
Is it $(R + I)/I \approx R/(R \cap I)$, the second isomorphism theorem? Then why isn't isomorphism mentioned?
On
This has nothing to do with integral extensions.
There is a ring homomorphism $\alpha:R\to S/I$ obtained by composing the inclusion $R\subset S$ with the quotient $S/I$.
The First Isomorphism Theorem tells us that $R/\ker{\alpha}$ is isomorphic to its image in $S/I$. So all that we need to do here is verify that $\ker{\alpha} = R\cap I$.
But this is obvious: an element of $R$ is mapped to zero in $S/I$ exactly if it's contained in $I$.
In general, if $f:A\to B$ is a ring homomorphism, then $A/\ker f\simeq\operatorname{Im}f\subset B$, so $A/\ker f$ is isomorphic to a subring of $B$.
Now use the composition $f:R\subset S\to S/I$ and notice that its kernel is $I\cap R$, so $R/I\cap R$ is isomorphic to a subring of $S$. By abuse of language one ca say that $R/I\cap R$ is a subring of $S$.