Bond length comparison between sulfur tetrafluoride and thionyl tetrafluoride
The simplest way to compare these two compounds is to compare sulfur’s oxidation state. In $\ce{SF4}$, sulfur is in the $(\mathrm{+IV})$ state. In $\ce{SOF4}$, it is $(\mathrm{+VI})$. The higher a compound’s oxidation state, the higher its formal positive charge and thus the more stable (i.e. lower in energy) its orbitals are. Orbitals reduced in energy means contracted orbitals which in turn leads to shorter bonds.

Other than that, the two compounds are pretty identical. Both feature two two-electron-two-centre $\ce{S-F}$ single bonds and one four-electron-three-centre $\ce{F\bond{...}S\bond{...}F}$ bond; the latter can be understood with the following resonance depiction:

$$\ce{F^-\ \ S^+-F &lt;-&gt; F-S^+\ \ F-}$$

Even the structures are comparable although the angles are slightly different.