You are not doing much wrong except for assigning improper oxidation number to oxygen of the peroxo group.
If the redox reaction “feels” wrong, you can check yourself using a table of standard reduction potentials [1, p. 5-79]:
$$
\begin{align}
\ce{H2\overset{-1}{O}_2 &<=> \overset{0}{O}_2 + 2 H+ + 2 e-} &\quad E^\circ_1 &= \pu{-0.695 V} \tag{1}\\
\ce{H\overset{+1}{Cl}O + H+ + 2 e- &<=> \overset{-1}{Cl}^- + H2O} &\quad E^\circ_2 &= \pu{+1.482 V} \tag{2}\\
\hline
\ce{H2\overset{-1}{O}_2 + H\overset{+1}{Cl}O &-> H3O+ + \overset{-1}{Cl}^- + \overset{0}{O}_2} &\quad E^\circ &= +\pu{0.787 V} \tag{3}
\end{align}
$$
Since resulting $E^\circ = \pu{0.787 V} > 0,$ free Gibbs energy $Δ_\mathrm{r}G^\circ = -nFE^\circ < 0,$ and the redox reaction where hydrogen peroxide is oxidized by chlorate(I) can be considered a thermodynamically favorable process.
Alternatively, you can use Latimer diagram, which comes in handy for disproportionation and synproportionation reactions.
Reference
- Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.