Understanding PDA and Equivalence of PDA and CFG
Your question is, in essence,


  Why does the proof that every context-free grammar can be converted to an equivalent PDA proceed in a particular way rather than in another way?


It's hard to answer such a question unless it gets more specific. For example, you can ask why the resulting PDA have to invoke nondeterminism. The answer is that some context-free languages cannot be accepted by a DPDA. Indeed, while this particular proof uses PDAs acting in a certain way, another proof might use PDAs acting in a different way.

One alternative such proof uses the Chomsky–Schützenberger representation theorem. The theorem states that every context-free language can be realized as
$$ h(D \cap R), $$
where $D$ is a Dyck language (the language of all correctly nested strings of parentheses of some fixed number of sorts), $R$ is a regular language, and $h$ is a homomorphism. This theorem, which can be proved directly using context-free grammars (see, for example, Context-free languages and pushdown automata by Autebert, Berstel and Boasson, from the Handbook of Formal Languages), allows to convert a context-free grammar to a PDA which is more like the one for $0^n 1^n$, along the following lines:


Start with a PDA for $D$. This is a PDA that pushes whenever encountering a left parenthesis, and pops whenever encountering a right parenthesis (checking that the two parentheses, the one on the stack and the one being read, have the same type).
Construct a DFA/NFA for $R$, and use the product construction to construct a PDA for $D \cap R$.
Construct a PDA for $h(D \cap R)$ by replacing each transition on $\sigma$ to a transition on $h(\sigma)$.


If you apply this construction to $0^n1^n$ (which you can realize as $D \cap 0^*1^*$, where $D$ is the Dyck language with a single type of left parenthesis $0$ and the corresponding right parenthesis $1$) then you get a PDA which is remarkably similar to the one you describe.