Will NaF + CaCO3 precipitate much CaF2?
Yes, the calcium ion could lead to precipitation. The solubility of $\ce{CaCO3}$ in distilled water is about 15 mg/L, which is about 0.15 mM calcium ion if there is no other source of carbonate. 

The solubility constant for $\ce{CaF2}$ is about $4\times 10^{-11}$, which means that we can only have 0.5 mM fluoride ions before precipitation will start. That's well below the solubility of NaF. 

However, the presence of additional carbonate could be used to reduce the calcium concentration. Similarly, any addition of acid will convert carbonate to bicarbonate and increase the maximum calcium concentration. 

UPDATE:
In the above answer, I assumed a basic familiarity with solubility products. For those not familiar with those, here's more detail.

The key quantitative measure of solubility of ionic compounds is the solubility product, usually indicated as $K_{sp}$, which is the product of the concentrations of the separate ions. 

For $\ce{CaCO3}$, we have $K_{sp}=[\ce{Ca^2+}][\ce{CO3^2-}]$. Reported values vary somewhat, but are typically around $2\times 10^{-8}$. 

Likewise, for $\ce{CaF2}$, we have $K_{sp}=[\ce{Ca^2+}][\ce{F-}][\ce{F-}]$, and the reported values are around $4\times 10^{-11}$. Thus, the condition for keeping fluoride in solution is 

$[\ce{Ca^2+}][\ce{F-}][\ce{F-}]&lt;4\times 10^{-11}$.

If $\ce{CaCO3}$ is dissolved in distilled water to maximum solubility, the calcium ion concentration is $\sqrt{K_{sp}}\approx 0.15$ mM, which is the basis of the calculation above.