The following expression can't take on a rational value for some real $h$: $$\frac{\sqrt{4 + h} -2}h$$
But this expression can:
$$\frac1{\sqrt{4 + h} + 2}$$
This is confusing me because the only factor that seems to have an impact on rationality/irrationality is the term $\sqrt{4+2}$, which appears in both expressions. So, what am I missing?
Can someone explain to me why one expression can take on a rational value and not the other?
Well neither of these expressions is rational, but presumably this appears in the context of computing limits such as: $$\lim_{h \to 0}\frac{\sqrt{4 + h} -2}{h}$$ which, by direct substitution, leads to the indeterminate form $\tfrac{0}{0}$. In this example, you can then rationalize the numerator as follows: $$\lim_{h \to 0}\frac{\sqrt{4 + h} -2}{h}=\lim_{h \to 0}\frac{\left(\sqrt{4 + h} -2\right)\color{blue}{\left(\sqrt{4 + h} +2\right)}}{h\color{blue}{\left(\sqrt{4 + h} +2\right)}}=\lim_{h \to 0}\frac1{\sqrt{4 + h} + 2}$$ and you arrive at the limit of your second expression. Although now the denominator isn't rational, you have rationalized the numerator and eliminated the 'problem', i.e. the indeterminate form $\tfrac{0}{0}$.
So coming back to your question:
It's safer to say that you have rationalized the numerator by replacing the first expression by the second. Note that this is not specific to the numerator since it also works the other way around; in the following limit, you could rationalize the denominator with the same trick: $$\lim_{h \to 0}\frac{h}{\sqrt{4 + h} -2}$$