I like to develop intuition with a specific universe. In this case, I take $x$ as having two values, say $0$ and $1$.
Then the first statement becomes $(\lnot P(0) \lor Q(0)) \lor (\lnot P(1) \lor Q(1))$.
The second statement becomes $(\lnot(P(0) \lor P(1)) \lor (Q(0) \lor Q(1))$.
It is straightforward to see that the assignment $P(0) = F, P(1)=T, Q(0) = F, Q(1) = F$ results in the first formula having the value $T$, while the second is $F$. Hence the formulas are not equivalent.
Rewriting the formulas as $\lnot P(0) \lor \lnot P(1) \lor Q(0) \lor Q(1)$ and $(\lnot P(0) \land \lnot P(1)) \lor Q(0) \lor Q(1)$ suggests that the second formula implies the first.
To prove this (loosely) we need only show that if the second formula is true, then so is the first.
The second formula can be true in two ways:
(1) $\exists x P(x)$ is false, in which case $\forall x \lnot P(x)$ is true, from which is follows that $\forall x (\lnot P(x) \lor Q(x))$ is true, from which we have $\forall x (P(x) \Rightarrow Q(x))$ and finally $\exists x (P(x) \Rightarrow Q(x))$ is true.
(2) $\exists x P(x)$ is true and $\exists x Q(x)$ is true. Since $\exists x Q(x)$ is true, it follows that $\exists x (\lnot P(x) \lor Q(x))$ is true, which is equivalent to $\exists x ( P(x) \Rightarrow Q(x))$.
Formalizing the proof needs more input from you...