Pumping lemma: why x in ∣xy∣ ≤ p?
You said:

If I understand correctly, $x$ and $z$ are basically anything on the 2
sides of the string $y^p$ that we're pumping and thus can be anything
in .

This is not true. The pumping lemma suggests that for every long enough word $w$  such that $w\in L$, there is a partition of $w$ into three words $w = xyz$ such that the three conditions of the lemma hold. You only know that such $x, y$ and $z$ exist, you do not know whether this is true for all $x, y, $ and $z$ that partition $w$. Thus, you cannot force $x$ or $z$ to be empty. In fact, this is far from being true. For example, consider the language of the regex $a b^*$. Here $y$ cannot contain the letter $a$, thus you cannot force $x$ to be the empty word.
You also said:

If $x$ is necessary, why is there no $|yz|\leq p$?

This condition is too strong as, along with the condition on $|xy|$, it suggests that every long enough word $w$ in the language, has to be of length at most $2p$, which is not correct.
Also, once you have the condition $|xy|\leq p$, there is no need to have a condition on $|yz|$. The pumping lemma conditions come up naturally from how a run of length at least $|Q|+1$ looks. In such long runs, two states have to repeat in the first $|Q|+1$ states of the run, and thus the condition on $|xy|$ arises. You can think about considering the last $|Q|+1$ states in a long run, and that might give you another variant of the pumping lemma (where $x$ and $z$ replace rules) but I don't see how this may be more useful or more appealing than the standard pumping lemma.