Why is the boiling point of water and ammonia so different?
If I interpret your edited question correctly, it now boils down to why is the difference between the boiling points of $\ce{HF}$ and $\ce{NH3}$ so large, even though they have almost the same molecular masses ($M_\mathrm{r}(\ce{NH3}) = 17, M_\mathrm{r}(\ce{HF}) = 20$).

Ammonia can self-ionise much like water can according to the following equation $(1)$. And much like water, the process is important to determine a $\mathrm{pH}$ value, but rather neglegible when it comes to determining attractive forces between ammonia molecules where ‘traditional’ hydrogen bonding predominates.

$$\ce{2 NH3 &lt;&lt;=&gt; NH4+ + NH2-}\tag{1}$$

On the other hand, self-ionisation in hydrogen fluoride, especially in the liquid and solid states, is much more important and leads to the two fragments shown in equation $(2)$.

$$\ce{3 HF &lt;=&gt; H-F-H+ + F-H-F-}\tag{2}$$

These two linear fragments both feature de facto four-electron-three-centre bonds and are thus much more prevalent than the corresponding ammonia-derived ones. (Indeed, something analogous to $\ce{FHF-}$ doesn’t exist for ammonia.) Since these also introduce ionic interactions into the equation, the intermolecular forces between $\ce{HF}$ molecules are much stronger than those between ammonia molecules contributing to $\ce{HF}$’s higher boiling point.