The idea is to make equivalence classes from pairs $(m,s)$ with $m\in M$ and $s\in S$ and denote the equivalence class of $(m,s)$ by $m/s$. We need to ensure that multiplying numerator and denominator by the same element of $S$ doesn't change the equivalence class, so $(mt)/(st)$ should be the same as $m/s$.
But when should $m/s=n/t$? It should be so when $mt=ns$, but it turns out that this is insufficient to ensure an equivalence relation: this is just a sufficient condition to put $(m,s)$ and $(n,t)$ in the same equivalence class. On the other hand, we should have
$$
\frac{m}{s}=\frac{mu}{su},\qquad \frac{n}{s}=\frac{nu}{tu}
$$
for every $u\in S$. It turns out that defining
$$
(m,s)\sim(n,t) \quad\text{if and only if}\quad mtu=nsu \text{ for some } u\in S
$$
makes $\sim$ into an equivalence relation. Defining
$$
\frac{m}{s}+\frac{n}{t}=\frac{mt+ns}{st},\qquad \frac{m}{s}\frac{r}{t}=\frac{mr}{st}
$$
does not depend on the representatives of the equivalence classes and makes $S^{-1}M$ (the quotient set) into a module over $S^{-1}R$ (with the similar definitions for the ring structure).
Clearly, for every $s\in S$, we need $0/s$ to be the zero element in $S^{-1}M$. By the very definition, then
$$
\frac{m}{s}=\frac{0}{t}
$$
if and only if $mtu=0su=0$, for some $u\in S$. But then we see that it's equivalent to say that $mu=0$, for some $u\in S$. One direction has been shown, as $tu\in S$; for the other direction
$$
\frac{m}{s}=\frac{mu}{su}=\frac{0}{su}
$$
is the zero element.
