Your proposed reaction is unlikely
That is because the combination $\ce{CaF2 + H2O}$ (under tap water conditions) is much more stable than $\ce{CaO + 2 HF}$. In other words, the electronic energy of the former is lower and the system preferably stays in this state. That does not mean, however, that the reaction does not occur! It just means that the concentration of the products is incredibly small and probably you would not be able to detect them at all. Because of this, chemists tend to say that it does not take place, although that is technically incorrect.
HF does occur, but in tiny concentrations.
I think there is a more likely way how HF would form. Let us look at the reactions that are necessary to have $\ce{HF}$ in the fluoridated water. The first step is the solvation of the solid fluoride ($\ce{NaF}$ or $\ce{CaF2}$) in water: $\ce{NaF2(s) <=> Na+(aq) + F-(aq)}$ or $\ce{CaF2(s) <=> Ca+(aq) + 2 F-(aq)}$, respectively. Sodium and calcium ions are stable in aqueous solution, so we can neglect them in the next steps. 870
Just imagine that they are still in the water, surrounded by water molecules. The second and last step is the fluoride ion reaction with water. According to Wikipedia, the reaction $\ce{F- + H2O <=> HF + OH-}$ has a $pK_b$ value of 10.8. This means: $$ pK_b = -lg(K_b) \Rightarrow K_b = 10^{-pK_b} $$ $$ K_b = \frac{[\ce{HF}] \cdot [\ce{OH-}]}{[\ce{F-}]} = 10^{-10.8}$$
So here we can see that in a solution containing only fluoride ions and inert counterions, the portion of hydrofluoric acid is incredibly low! As mentioned above, you would probably not be able to detect it.
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Cunningham's Law worked here for me in the comments.
Generally speaking, if you have a chain of reactions that amounts to nothing happening, it just doesn't happen in the first place. It is as simple as that.
That is a severe misunderstanding of chemical reaction dynamics. In fact, the exact opposite is true: If you have a chain of reactions, all of them happen! The question is only to which extent, which I tried to answer above. See here and here for more details. 979