I know that a universal quantifier can be distributed over conjunction and not disjunction, but I'm having a hard time wrap my head around it. Why is this the case? Is there an example of a statement that would demonstrate this principle?
Matt E, in his comment, gives two statements. He shows that a universal quantifier cannot be distributed over disjunction, because if it could, a false statement would follow logically from a true one. The statement "A universal quantifier can be distributed over disjunction" is itself a universal statement, and requires only one counterexample to disprove it.
"A universal quantifier can be distributed over conjunction" is also a universal statement. It cannot be proved by giving examples. There might be some counterexample we didn't think of. So how do we know it's true? -- Good question!
Intuitively, the reason that universal quantification can be distributed over conjunction is that universal quantification can already be viewed as conjunction: $(\forall x)\Phi(x)$ can be viwed as the conjunction of $\Phi(d)$ taken over every element $d$ of the domain. Viewed this way, $(\forall x)(\Phi(x) \land \Psi(x))$ and $(\forall x)\Phi(x) \land (\forall x)\Psi(x)$ are equivalent because they both represent a giant conjunction of every instance of $\Phi(d)$ along with every instance of $\Psi(d)$. This can be made more precise by looking at Tarski's schema for truth in a structure.
For the same reason, existential quantification can be distributed over disjunction, because $(\exists x)\Phi(x)$ can be viewed as the disjunction of every possible substitution instance of $\Phi$.
In very old literature, people actually used $\bigwedge_x$ for $\forall$ and $\bigvee_x$ for $\exists$, for this reason.
Now, continuing this informal viewpoint, the reason that $\forall$ cannot be distributed over disjunction is that a certain distributive rule does not hold: $$ \bigwedge_x\left(\Phi(x) \lor \Psi(x)\right) $$ is not in general the same as $$ (\bigwedge_x \Phi(x)) \lor (\bigwedge_x \Psi(x)) $$ Indeed, this rule already fails when there are just two elements in the domain, because the propositional formula $(P \lor Q) \land (R \lor S)$ is not equivalent to $(P \land R) \lor (Q \land S)$.