I am not really sure how to phrase it correctly, but what I am interested in is the following:
... lets say I have a proposition of this form:
$$(\forall x : \text{statement 1}) \iff \text{statement 2}$$
Is this the equivalent of:
$$\forall x : (\text{statement 1} \iff \text{statement 2})$$
On
No.
There is an equivalence regarding the one-way conditional (where $Q$ makes no reference to $x$):
$$\forall x (Q \to P(x)) \Leftrightarrow Q \to \forall x P(x)$$
OK, but note that given that equivalence, you have:
$$\forall x (P(x) \to Q) \Leftrightarrow$$
$$\forall x (\neg Q \to \neg P(x)) \Leftrightarrow$$
$$\neg Q \to \forall x \neg P(x) \Leftrightarrow$$
$$\neg Q \to \neg \exists x P(x)) \Leftrightarrow$$
$$\exists x P(x) \to Q$$
So:
$$\forall x (P(x) \to Q) \Leftrightarrow \exists x P(x) \to Q$$
So: when the quantifier is in the antecedent, the quantifier changes its 'sign' when pulling it 'inside' or 'outside' the conditional.
And, because of the fact that this does not happen when the quantifier is in the consequent of the conditional, there is no nice equivalence principle for a quantifier in connection with a bi-conditional.
What you have instead is:
$$\forall x (P(x) \leftrightarrow Q) \Leftrightarrow$$
$$\forall x ((P(x) \to Q) \land (Q \to P(x)))\Leftrightarrow$$
$$\forall x (P(x) \to Q) \land \forall x (Q \to P(x))\Leftrightarrow$$
$$(\exists x P(x) \to Q) \land (Q \to \forall x P(x))$$
No, for example consider $$ \tag{a} (\forall x: x=0) \iff 1=0 $$ versus $$ \tag{b} \forall x:(x=0\iff 1=0) $$ Formula (a) is true because both sides of the $\iff$ have the same truth value, namely false.
Formula (b) is false, because setting $x=0$ makes the inner claim fail: $$ 0=0 \iff 1=0 \qquad\text{is definitely not true}$$
What is the case is that $$[\forall x: (p(x) \Leftrightarrow q)] \implies [(\forall x: p(x))\Leftrightarrow q] $$ when $x$ does not appear free in $q$. But only that direction holds in general.