Why is tin(II) fluoride more effective in turning apatite into fluorapatite?
Found an answer to this question in a book called 'Pediatric Dentistry'. 

Apparently, when Stannous Fluoride reacts with dental enamel (hydroxyapatite), both the fluoride and the tin react to form Stannous Trifluorophosphate $(\ce{Sn3F3PO4})$ -- This is highly resistant to acid attack.

At high concentrations of stannous fluoride (I don't know what counts as "high", the book doesn't say), apparently the following compound is formed as well: Trifluorostannate -- This is also highly acid resistant.

In addition, Calcium Fluoride is also produced, which then further reacts with hydroxyapatite forming Fluorhydroxyapatite (which is also rather acid resistant, although it seems that the stannous trifluorophosphate is more resistant.)

Sodium fluoride only produces Fluorhydroxyapatite as its reaction. 

It appears that Stannous Fluoride is a superior agent because it produces 1 or 2 additional compounds (that become part of the surface enamel) that are more acid resistant than just the one that Sodium Fluoride produces. These other compounds (which become a part of the surface of the tooth) also have an anti-microbial effect, thus inhibiting the growth of the very bacteria which are directly responsible for producing the majority of the acid found in dental decay.