As the Rotman's, or Voloch's, constructions of Dubuque's answers show, you can't recover exactly "the usual definition" of the localization of a ring from its universal property. That is, a universal property won't tell you, in general, how to construct an object that verifies it (as you can see, there are at least, two constructions for the localization, both verifying the universal property: how are you going to choose among them?). What a universal property will tell you is that all the guys verifying it are necessarily isomorphic: for instance, the construction in Vakil's notes, and Rotman's are necessarily isomorphic.
Precisely, let's stay with rings and let $\gamma' : A \longrightarrow B$ be some guy that satisfies the universal property of the localization. Let $\gamma: A \longrightarrow S^{-1}A$ denote the construction of the localization explained in Vakil's notes, 2.3.3. Then, since both $S^{-1}A$ and $B$ satisfy the universal property, you'll have morphisms $\widetilde{\gamma} : S^{-1}A \longrightarrow B$ and $\widetilde{\gamma'}: B \longrightarrow S^{-1}A$ such that $\widetilde{\gamma'}\gamma = \gamma'$ and $\widetilde{\gamma}\gamma' = \gamma$. Let's show that both $\widetilde{\gamma}$ and $\widetilde{\gamma'}$ are inverses of each other. For instance, $\widetilde{\gamma'} \widetilde{\gamma} = \mathrm{id}_B$ because
$$
(\widetilde{\gamma'} \widetilde{\gamma})\gamma' = \widetilde{\gamma'}\gamma = \gamma' \ .
$$
But $\mathrm{id}_B$ verifies the same identity clearly:
$$
\mathrm{id}_B \gamma'= \gamma' \ .
$$
Hence, because of the uniqueness of the universal property, you must have
$$
\widetilde{\gamma'} \widetilde{\gamma} = \mathrm{id}_B \ .
$$
Analogously,
$$
\widetilde{\gamma}\widetilde{\gamma'} = \mathrm{id}_{S^{-1}A} \ .
$$
What you can (and must) do is to verify that the usual definition (particular construction) of the localization verifies the universal property. That is, given some morphism of rings $\varphi : A \longrightarrow B$ such that $\varphi s \in B$ is a unit for every $s \in S$, then there is a unique morphism of rings $\widetilde{\varphi} : S^{-1}A \longrightarrow B$ such that $\widetilde{\varphi}\gamma = \varphi$.
Hint: you can easily check that
$$
\widetilde{\varphi} \left( \frac{a}{s} \right) = \frac{\varphi (a)}{\varphi (s)}
$$
is well-defined (that is, if $\frac{a}{s} = \frac{b}{t}$, then $\widetilde{\varphi}\left( \frac{a}{s} \right) = \widetilde{\varphi}\left( \frac{b}{t} \right) $), is a morphism of rings, verifies $\widetilde{\varphi} \gamma = \varphi$ and is unique (that is, if you had another $\psi : S^{-1}A \longrightarrow B$ such that $\psi\gamma = \varphi$, then $\psi = \widetilde{\varphi}$).