According to Vakil's Rising Sea, when we have localized ring, then two fractions $\frac{a}{b}$ , $ \frac{c}{d}$ are only equal iff there exists an $s$ in the ring you quotiented by such that:
$$ s(ad-bc)=0$$
My question is, why do we have an $s$ here? Why not just say, that the two are equal when ad=bc?
Consider $R = \mathbb{Z} / 6 \mathbb{Z}$, and let $S$ be the multiplicatively closed set generated by $2$. Then we should have $\frac{3}{1} = \frac{0}{2}$, since $3 \cdot 2 - 0 \cdot 1 = 0$. And we should also have $\frac{0}{2} = \frac{0}{1}$, since $0 \cdot 1 - 0 \cdot 2 = 0$. So $\frac{3}{1} = \frac{0}{1}$. But we do not have $3 \cdot 1 - 0 \cdot 1 = 0$. So we run into serious issues if we don’t have the $s$, since the resulting equivalence relation isn’t necessarily transitive.
Note that a notable case where we don’t need the $s$ is when $S$ doesn’t contain any factor of $0$. In particular, if $R$ is an integral domain and $0 \notin S$, then $\frac{a}{b} = \frac{c}{d}$ if and only if $ad - cb = 0$.