Does brewing tea with hard water reduce the amount of bioavailable fluorine?
$\ce{CaF2}$ has a relatively low solubility in water, about 15 mg/L, with a $K_{sp}$ of about $4\times 10^{-11}$.  According to the paper you linked, the concentrations of fluoride in tea are around 5 mg/L, which corresponds to about $1.3\times 10^{-4}$ M. Thus, the concentration of $\ce{Ca^2+}$ required to precipitate the fluoride out is about 2.5 mM, which corresponds to 0.1 g/L. Hard water is defined as having a calcium concentration above about 0.6 mM, with very hard water being 1.8 mM calcium. Thus, it is unlikely that there would be high enough concentrations for $\ce{CaF2}$ to precipitate out of solution.

Because of the large solvation energy of fluoride ion, the molecules in solution are likely to be nearly completely ionized. That is, the fluoride anions and the calcium cation will each have their own solvation shell of water molecules. That means that they will not interfere with each other until the concentrations are high enough that solid $\ce{CaF2}$ begins to precipitate. 

As long as the concentrations are below the solubility limit, the presence of calcium ions will therefore not affect the availability of the fluoride ions. 

Furthermore, since HF is a weak acid, in the acidic environment of the stomach some of the fluoride ions will be protonated to HF, decreasing the concentration of free fluoride and increasing the apparent solubility of $\ce{CaF2}$.