Caclulating the Solubility of Fluorite in River Water
the ppm of fluorite in pure water (15.6ppm)


the ppm of fluorite in pure water is zero!


  $K_\mathrm{sp}=[\ce{Ca_{original}} + \ce{Ca_{added}}][2\ce{Ca_{added}}]^2 = 4[\ce{Ca_{original}} + \ce{Ca_{added}}][\ce{Ca_{added}}]^2$


Yes, this is correct and can be solved, you say you know $[\ce{Ca_{original}}]$ is $\pu{0.375E-3M}$ and  $K_\mathrm{sp}$ is $10^{-10.5}$ so there is one equation and one unknown, just a matter of solving a cubic equation, which can be solved exactly or by approximate methods.