The fraction ring (localization) $\rm\,S^{-1} R\,$ is, conceptually, the universal way of adjoining inverses of $\rm\,S\,$ to $\rm\,R.\,$ The simplest way to construct it is $\rm\,S^{-1} R = R[x_i]/(s_i x_i - 1).\,$ This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (avoiding the many tedious verifications always "left for the reader" in the more commonly presented pair approach). For details of this folklore see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, or Voloch, Rings of fractions the hard way.
Likely Voloch's title is a joke - since said presentation-based method is by far the easiest approach. In fact both Rotman's and Voloch's expositions can be simplified. Namely, the only nonobvious step in this approach is computing the kernel of $\rm\, R\to S^{-1} R,\,$ for which there is a nice trick:
$\quad \begin{eqnarray}\rm n = deg\, f\quad and\quad r &=&\rm (1\!-\!sx)\,f(x) &\Rightarrow&\ \rm f(0) = r\qquad\,\ \ \ via\ \ coef\ x^0 \\
\rm\Rightarrow\ (1\!+\!sx\!+\dots+\!(sx)^n)\, r &=&\rm (1\!-\!(sx)^{n+1})\, f(x) &\Rightarrow&\ \rm f(0)\,s^{n+1}\! = 0\quad via\ \ coef\ x^{n+1} \\
& & &\Rightarrow&\ \rm\quad r\ s^{n+1} = 0
\end{eqnarray}$
Therefore, if $\rm\,s\,$ is not a zero-divisor, then $\rm\,r = 0,\,$ so $\rm\, R\to S^{-1} R\,$ is an injection.
For cultural background, for an outstanding introduction to universal ideas see 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: Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002.
Best Answer
I don't know how your book defines the ring of fractions, but the right (i.e. the most general as possible) way of doing things with rings of fractions is this :
If $M$ is an $R$-module and $1 \in D \subseteq R$ is a subset of $R$ which is multiplicatively closed, then we define $D^{-1} R$ as follows. Define the following equivalence relation over $D \times M$ : $(d_1, m_1) \sim (d_2, m_2)$ if there exists $d \in D$ such that $d(d_1 m_2 - d_2 m_1) = 0$. You can check that this defines an equivalence class over $D \times M$ (the details are okay to work out, nothing hard), and in the case where $M = R$ and $R$ is an integral domain, you can get rid of the condition 'there exists a $d \in D$ such that' because it is not necessary.
Defining the addition as usual over $D^{-1} M$, this makes $D^{-1}M$ into an abelian group. In particular, since $R$ is an $R$-module over itself, we can define $D^{-1}R$. Considering the particular case of $D^{-1}R$ alone first, we can define multiplication as $\frac{r_1}{d_1} \frac{r_2}{d_2} = \frac{r_1 r_2}{d_1 d_2}$. This makes $D^{-1}R$ into a ring, so now we can say that the scalar multiplication $\frac{r}{d} \frac{m}{d'} = \frac{rm}{dd'}$ makes $D^{-1}M$ into a $D^{-1}R$-module.
(I used the letter $D$ because it stands for denominators. The letter $S$ is probably just used because of the alphabet...)
In this kind of generality, if you want to make $D^{-1}R$ into an integral domain, you need to make some assumptions on $D$. For instance,
Note that the set $D$ of all non-zerodivisors is a multiplicatively closed subset of $R$ which contains $1$. This means that if $\frac a1 = \frac b1$, there exists $d \in D$ such that $d(a-b) = 0$, but $d$ is not a zero-divisor, then $a-b =0$ and $a=b$, hence the remark that $f$ is a monomorphism in this case. If $D$ contains a zero divisor, you can have some fraction $\frac r1 = \frac 01$ without having $r=0$, because this equation only means that there exists $d \in D$ such that $rd = 0$. The equivalence class of $\frac 01$ contains precisely those $r \in R$ that are zero divisors.
In an integral domain, there are no zero-divisors, so by the above remark the map $f$ is an embedding of $R$ into its quotient field $D^{-1}R$ (where $D$ is the set of non-zero elements, which is also the set of non-zero divisors in this case). Note that $D^{-1}R$ is a field because a fraction $\frac rd$ is never equivalent to $\frac 0{d'}$ for some $d'$, hence we can take its inverse to be $\frac dr$.
Hope that helps! Feel free to ask any questions about the details.