Generalization bound (single hypothesis) in "Foundations of Machine Learning"
You are right. The relaxed inequality 
$$R(h) \le  \hat{R}_S(h)+ \epsilon.$$
can be replaced with the complete inequality
$$\left |\hat{R}_S(h) - R(h) \right| \le \epsilon.$$
Actually, authors use this complete inequality for the follow up examples in the book. Again in Theorem $2.13$, they write the relaxed inequality, but prove for the complete inequality.

We could say that the relaxed inequality is written for the sake of readability and/or convention.

On the relation of inequalities

Let us denote:
$$A:=\hat{R}_S(h) - R(h) \le \epsilon$$
$$B:=\hat{R}_S(h) - R(h) \ge -\epsilon$$
thus,
$$\left| \hat{R}_S(h) - R(h) \right| \le \epsilon = A \text{ and } B$$
Equation $(2.16)$ states:
$$\begin{align*}
&amp; {\Bbb P}(\left| \hat{R}_S(h) - R(h) \right| \ge \epsilon) \le \delta \\
&amp; \Rightarrow {\Bbb P}(\left| \hat{R}_S(h) - R(h) \right| \le \epsilon) \ge 1 - \delta \\
&amp; \Rightarrow {\Bbb P}(A \text{ and } B) \ge 1 - \delta \\
\end{align*}$$
knowing that ${\Bbb P}(B) \ge {\Bbb P}(A \text{ and } B)$,
$$\begin{align*}
&amp; {\Bbb P}(B) \ge {\Bbb P}(A \text{ and } B) \ge 1 - \delta \\
&amp; \Rightarrow {\Bbb P}(\hat{R}_S(h) - R(h) \ge -\epsilon) \ge 1 - \delta
\end{align*}$$
which is equivalent to 
$$R(h) \le \hat{R}_S(h) + \epsilon$$
with probability at least $1-\delta$, i.e. equation $(2.17)$.