What Definition 3.1(a) wants to convey is the following: say that you start with an axiom system $\Phi$ and you want have a way of "picking-up" $n$-tuples of elements in models $M$ of $\Phi$ which belong to a subset of $M^n$ satisfying some first order property given by an $S$-formula $\varphi_P(v_0, \dots, v_{n-1})$. To do so, one introduces a new $n$-ary relation symbol $P$ and extends the axiom system $\Phi$ to a new axiom system $\Phi \cup \{\delta_P\}$ in the new symbol set $S' = S \cup \{P\}$, where $$\delta_P := \forall v_0, .... \forall v_{n-1} (P v_0 ... n_{n-1} \leftrightarrow \varphi_P(v_0, ... , v_{n-1}));$$
the $S'$-sentence $\delta_P$ fixes the interpretation of the new symbol $P$ in models of $\Phi \cup \{\delta_p\}$. Now, the Theorem on Definitions (3.2) ensures us that we're not doing anything silly (namely, that both $\Phi$ and $\Phi \cup \{\delta_P\}$ prove the same theorems, so $\Phi \cup \{\delta_P\}$ is a conservative extension of $\Phi$) and that we really didn't had to add any new symbol to $S$ in order to pick-up those $n$-tuples of elements (i.e. $\varphi_P$ plays exactly the same role in models of $\Phi$ as $P$ does in models of $\Phi \cup \{\delta_P\}$).
The point is that to extend the symbol set $S$ by a new relation symbol $P$, there is no further condition which our axiom system $\Phi$ needs to satisfy, and that is why it seems as if $\Phi$ is not used in the definition of an $S$-definition for $P$. This is not however the case for the extension by definitions for constant symbols and function symbols. For the case of constant symbols one needs to ensure that the $S$-formula $\varphi_c$ defining the new constant symbol $c$ uniquely determines the interpretation of $c$ in models of $\Phi$, i.e. we need to impose the condition that $\Phi \models \exists^{=1}v_0\varphi_c(v_0)$. Definition 3.1(b) for the $S$-definition of a function symbol follows a similar idea.
