Primitive recursive functions and unbounded quantifiers
If $P(t,x_1,\dots,x_n) \in \zeta$, then clearly $(\forall t)_{\le 0}P(t,x_1,\dots,x_n) = P(0,x_1,\dots,x_n) \in \zeta$.

Moreover, if for some $y$, we have $(\forall t)_{\le y}P(t,x_1,\dots,x_n) \in \zeta$, then it can easily be shown that $(\forall t)_{\le y+1}P(t,x_1,\dots,x_n) = P(y+1,x_1,\dots,x_n) \land (\forall t)_{\le y}P(t,x_1,\dots,x_n) \in \zeta$, noting that taking the logical and is a primitive recursive operation.

By induction, we have that for any $y$,  $(\forall t)_{\le y}P(t,x_1,\dots,x_n) \in \zeta$. Things work similarly for $(\exists t)_{\le y}$. However, this reasoning doesn't extend to unbounded quantifiers, and in the general case, $\forall t P(t,x_1,\dots,x_n) \notin \zeta$.

If you think of PRC predicates as things that can be checked by a computer in finite time, the underlying meaning is that


if it takes finite time to check whether a given proposition is true for any given value, then it also takes finite time to check whether it's true for any/all values in a finite set (at most the sum of the finite times taken to check for each element in the set)
on the other hand, it's not necessarily possible to check that it's true for any/all values in an infinite set. Naively checking for $0$, then $1$, then $2$, etc..., could take forever.


The answer to your second question is proposition 4. Indeed, in proposition 3, the universal quantifier over $x$ is bounded by the free variable $y$, which is a finite value. In other words, proposition 3 could be rewritten as $(\forall x)_{\le y}P(x)$.