The formula for stannous fluoride is SnF4.
The tin cation has a charge of 4+ and a fluorine anion has a charge of 1-. Using the crisscross method, we move the charge of the tin cation to the subscript position of the fluorine anion and the charge of the fluorine anion to the subscript position of the tin cation. So, we get Sn1F4. The ratio one-to-four cannot be reduced further. Since the subscript of Sn is 1, we can drop it. So the formula of stannous fluoride becomes SnF4.