There's something I am missing here and I don't know what it is.
I understand that the Stone-Čech compactification of $X$ satisfies the property that for every continuous map $f: X \rightarrow K$ where $K$ is compact $T_2$, there is a unique continuous function $\beta f: \beta X \rightarrow K$ that extends $f$.
I don't understand why $K$ needs to be compact, or why this only happens for $\beta X$. I mean, given any compactification $Y$ of $X$, we can see $X$ as a dense subset of $Y$. So if $f$ can be extended to some $\tilde{f}: Y \rightarrow K$ continuously, it is clear that $\tilde{f}$ is unique, $X$ is dense in $Y$. So is the EXISTENCE of such $\tilde{f}$ what makes the Stone-Čech compactification special?
tl;dr
In short, if $Y$ is a compactification of $X$ different than $\beta X$, there exists a compact $K$ and a continuous function $f: X \rightarrow K$ that cannot be extended continuously to a function $\tilde{f}:Y \rightarrow K$?
Yes, precisely. Take $K = \beta X$. If the canonical embedding $f \colon X \to \beta X$ has a continuous extension to a compactification $Y$ of $X$, then $\tilde{f}(Y)$ is a dense (since it contains $f(X)$) compact (since $Y$ is compact) subset of $\beta(X)$, so $\tilde{f}(Y) = \beta X$. On the other hand, from the inclusion $g \colon X \to Y$, we have $\beta g\colon \beta X \to Y$. By the denseness of $X$, we must have $\beta g\circ \tilde{f} = \operatorname{id}_Y$ and $\tilde{f}\circ \beta g = \operatorname{id}_{\beta X}$.
So $\beta X$ is - up to isomorphism - the only compactification such that all continuous $f\colon X \to K$ with a compact $K$ can be extended. The inclusion in the Stone-Čech compactification cannot be extended to any other (non-isomorphic) compactification.