I think there are a variety of qualitative ways of looking at this:
- Perhaps the most obvious is that $\ce{H2O}$ can form a greater number of hydrogen bonds due to having an equal numbers of hydrogen bond acceptors and donors. Each of the hydrogen atoms can be hydrogen bond acceptors; each of the lone pairs on the oxygen can be donors. In $\ce{HF}$ however there is only one hydrogen bond acceptor and theoretically three hydrogen bond donors. This imbalanced ratio inevitably leaves some acceptors without a donor. Imagine this as two dance parties; one party has two men and two women but the other has one man and three women. Obviously fewer bonds can be successfully formed in the latter, assuming that everything is monogamous.
- As mentioned by the other poster, both O and F are very electronegative elements. F however is more EN than O, which means that F stabilizes electrons better than O. As a result, electrons localized on F are weaker hydrogen bond donors; they are already fairly well stabilized. Remember, electrons "want" to be stabilized by nuclei. If they already are fairly well stabilized then they will "feel" less of a need to be associated (and therefore stabilized) with (by) other nuclei. This suggests that the hydrogen bonding present among $\ce{HF}$ molecules might be weaker than the hydrogen bonds present among $\ce{H2O}$ molecules.
Experimental results are the gold standard, of course, as opposed to qualitative reasoning, and qualitative reasoning here can also lead us in the opposite direction; one might argue that by having three lone pairs, F has a lot on its plate, so to speak; it might be able to stabilize one lone pair of electrons very well but three lone pairs is a bigger ordeal, and perhaps big enough that the hydrogen bonds among $\ce{HF}$ molecules are stronger than those among water molecules. Again, this is all qualitative, but this is the sort of reasoning introductory chemistry teachers desire.
If we limit our thinking to just electrostatics then we might surmise that $\ce{HF}$ should have the stronger hydrogen bonds because F is more electron withdrawing and therefore hydrogen should be more positively polarized in $\ce{HF}$ as opposed to hydrogen in water. However, hydrogen bonding is more than just electrostatics. Hydrogen bonding actually has a covalent component; this, however is usually ignored by introductory treatments of chemistry. The bond angle of elements involved in a hydrogen bond is critical. The closer the elements involved in a hydrogen bond are to 180 degrees, the stronger the bond (this specific angle is the case with hydrogen bonding in water; not necessarily other molecules). If hydrogen bonding were purely electrostatic then this would not be the case; angles would not matter - only distance would.
Other issues must be explored as well, such as the number of viable hydrogen bonds and the electron-donating/releasing tendencies of the involved elements.
After scouring the web some common "explanations" that would not explain why water has a higher boiling point than $\ce{HF}$ would be:
water can form 4 per molecule while HF can only form 2.
Incorrect because if we only look at hydrogen bonding as involving the polarity of atoms, then how can water form four hydrogen bonds per molecule? It has two positively charged hydrogens, and a negatively charged oxygen. Seems like it should only form three hydrogen bonds. One must understand that lone pairs can each be hydrogen bond donors.