Why are fluorides of transition metals unstable in low oxidation states?
You can take the example of $\ce{CuF}$ which is not known.

Reason: $\ce F$ is such an electronegative element that it will always oxidise $ \ce{Cu}$ to $\ce{Cu^2+}$ and not $\ce{Cu^1+}$. Hence, whenever $\ce{Cu}$ reacts with flourine, copper(II) flouride is formed. The following reaction illustrates this:

$$\ce{Cu + F2 -&gt; CuF2}$$

$\ce{CuF2}$ loses fluorine at temperatures above $\pu{950 °C}$.  $\ce{CuF}$ will be formed initially and as your question says, it is highly unstable so it will be transformed to some other stuff and the reactions are:

$$\ce{2CuF2  -&gt; 2CuF (unstable) + F2}$$

$$\ce{2CuF  -&gt; CuF2 + Cu}$$

Similarly, you can illustrate this for $\ce{VF2}$.