In this context the fraction $\frac{\binom42}{\binom{52}5}$ has no very natural interpretation. The original fraction $\frac{\binom42\binom{48}3}{\binom{52}5}$ is another story: the numerator is the number of $5$-card hands containing exactly two kings, and the denominator is the number of $5$-card hands, so the fraction is the probability of being dealt a hand containing exactly two kings.
Just as there are $\binom42\binom{48}3$ hands with exactly two kings, there are $\binom43\binom{48}2$ hands with exactly three kings and $\binom44\binom{48}1$ hands with exactly four kings. Thus, the number of hands with at least two kings is $\binom42\binom{48}3+\binom43\binom{48}2+\binom44\binom{48}1\;,$ and the probability of being dealt such a hand is $\frac{\binom42\binom{48}3+\binom43\binom{48}2+\binom44\binom{48}1}{\binom{52}5}\;.\tag{1}$
Note: The number of hands with no kings is $\binom40\binom{48}5$, and the number with exactly one king is $\binom41\binom{48}4$, so the number with at most one king is $\binom40\binom{48}5+\binom41\binom{48}4\;.$ Thus, the number of hands with at least two kings is $\binom{52}5-\left(\binom40\binom{48}5+\binom41\binom{48}4\right)\;,$ the total number of possible hands minus the number having fewer than two kings. Thus, we could also have computed the probability in $(1)$ as
$\frac{\binom{52}5-\left(\binom40\binom{48}5+\binom41\binom{48}4\right)}{\binom{52}5}=1-\frac{\binom40\binom{48}5+\binom41\binom{48}4}{\binom{52}5}\;.$ This has a perfectly good intuitive significance: it’s $1$ minus the probability of getting a hand with fewer than two kings.