Oxidation state of fluorine in HOF
There is no anomaly in here. Fluorine’s oxidation state in $\ce{HOF}$ is $\mathrm{-I}$ as the theory says. Hydrogen’s is $\mathrm{+I}$. This leaves oxygen with an oxidation state of $\mathrm{\pm 0}$.

That final fact is where the oxidative power comes from. Oxygen, the second-most electronegative element in the periodic table, has the same oxidation state as in molecular form. It is also bonded to an even more electronegative fluorine. Even in hydrogen peroxide oxygen would me more reduced. Thus, any reaction of $\ce{HOF}$ involves oxygen gaining electrons, not fluorine.

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