How would one compare the magnitude of covalent character between SnCl4 and SnF2 using Fajan's Rules?
Fajan's rules for the prediction of bond character can be summarized in two points:


Large cation with low positive charge + small anion $\rightarrow$ ionic
Small cation with high positive charge + large anion $\rightarrow$ covalent


In $\ce{SnF2}$, we have a large cation with low positive charge. $\ce{Sn^2+}$ is larger than $\ce{Sn^4+}$ because it has 2 more electrons in the $5s$ orbital. The anion $\ce{F^-}$ is smaller than its heavier group homolog $\ce{Cl^-}$. For this combination large cation + small anion, the bond character is predicted to be more ionic and less covalent.

On the other hand, $\ce{Sn^4+}$ is a small cation with high positive charge, and $\ce{Cl^-}$ is the large anion, as already stated. Therefore, according to the rules, $\ce{SnCl4}$ is the tin halide with more covalent bonds, and indeed, tin(IV) chloride consists of tetrahedral $\ce{SnCl4}$ molecules in the solid, liquid and gaseous state.

However, the predictions made with these rules do not always agree with the real bonding situation, and $\ce{SnF2}$ is an example. There is experimental evidence for covalent $\ce{Sn-F}$ bonds in tin(II) fluoride (references: 1, 2).