Learning to think categorically (localization of rings and modules)

In my opinion, the most enlightening (and the simplest) way to present the universal construction of localizations (and fractions) is to use instead of the pair construction the natural presentation in terms of generators and relations. This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (and to avoid the many tedious verifications required in the pair approach). Moreover, this approach is much more conceptual. Indeed, the pairs in the pair construction are nothing but normal forms for the polynomial terms in the presentation based approach. For details of this approach see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, and Voloch's: Rings of fractions the hard way. Note: presumably Voloch's title is a joke - since the presentation based approach is actually the easiest way - in fact both Rotman's and Voloch's expositions can be simplified.

If you're just beginning to understand universal constructions then I highly recommend that you peruse the beautiful exposition in George Bergman's An Invitation to General Algebra and Universal Constructions.

You might also find illuminating Paul Cohn's historical article Localization in general rings, a historical survey - as well as other papers in that volume [1].

[1] Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002

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 $\psi = \widetilde{\varphi}$).