How would one compare the magnitude of covalent character between SnCl4 and SnF2 using Fajan's Rules? It is easy to compare two ionic compounds when one of the ions is same. However, how do we compare two compounds if one of the ions is the same element but just has different charge and the other ion is completely different?
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).