I’m struggling to get an intuition on the non-existence of countably saturated models for complete theories with uncountably many types... I know many people on here seem to abhor the “why?” question, but I can’t help but ask why is this the case, in order to fully understand the issue.
Below is an example in an infinite language; it's not hard to modify it to work in a finite language.
Actually, barring further context there's a serious typo in the text. All complete theories have countably saturated models; what's true is that we don't always have countable countably saturated models.
Interestingly, the existence of fully saturated models of any cardinality of arbitrary complete theories is independent of the usual axioms of set theory!
For $i\in\mathbb{N}$ let $U_i$ be a unary relation symbol. For $A,B$ disjoint finite sets of natural numbers, let $$\psi_{A,B}\equiv\exists x[\bigwedge_{a\in A}U_a(x)]\wedge[\bigwedge_{b\in B}\neg U_b(x)].$$ Let $$T=\{\psi_{A,B}: A,B\mbox{ disjoint finite sets of naturals}\}.$$
Intuitively, a model of $T$ amounts to a bunch of subsets of $\mathbb{N}$, possibly with repetition (and conversely, a multiset of subsets of $\mathbb{N}$ yields a model of $T$ precisely when it is appropriately "dense" in $\mathcal{P}(\mathbb{N})$).
If $M\models T$ is countably saturated, then for each $S\subseteq\mathbb{N}$ there must be some $c\in M$ such that for all $i$ we have $M\models U_i(c)$ iff $i\in S$. But this means that no model of $T$ with cardinality $<2^{\aleph_0}$ is countably saturated.
(It's not hard to modify the above to use only a finite language.)
In general, the point is that a structure's cardinality can't be smaller than the number of distinct types realized in it (trivially), and reasonably saturated models have to realize lots of types.