Another way to think about this comes as to see what quantifiers look in the context of classical propositional logic which has truth set {0, 1}. Since we only have two truth values here, if I say "for all p, p" I've basically said "p is false, and p is true." So, we can interpret, at least here, "for all" as meaning a conjunction "^". So, for any such formula p, $\forall p $(p) means (0^1), and $\forall p$ ((p v q) ^ q) means
(((0 v q) ^q)^((1 v q)^q)).
"There exists" means "for at least one", which we can interpret as meaning a disjunction "v" in this context. So, $\exists p$ means (0 v 1), and $\exists p$ ((p v q) ^ q) means (((0 v q) ^q)v((1 v q)^q)).
Now let us see how the quantifiers behave when we switch them.
$\forall q $ $\exists p$ ((pvq)^q) becomes
$\forall q $ (((0 v q) ^q)v((1 v q)^q))) which becomes
((((0 v 0) ^0)v((1 v 0)^0))^(((0 v 1)^1)v((1 v 1)^1)))
On the other hand
$\exists p$ $\forall q$ ((p v q)^q) becomes
$\exists p$ (((p v 0)^0)^((p v 1)^1)) which becomes
((((0 v 0)^0)^((0 v 1)^1))v(((1 v 0)^0)^((1 v 1)^1)))
If you think of "for all" as indicating a conjunction for 2-element sets as the above suggests one might do, then for sets with more than 3 elements, and sets with an infinity of elements, then "for all" will indicate an extended conjunction, and similarly you'll have "there exists" as an extended disjunction. What do you do with 1-element sets though under this interpretation? Simple, all quantifiers effectively become meaningless, and the distinction between the universal and existential quantifiers breaks down. So, you just write the element of the set in any formula and you can forget about quantifiers here.