How to apply oxidation numbers in the reaction to form hydrogen peroxide from a superoxide?
On the reactant side, the superoxide has a $-1$ charge per oxygen atom.


No, the superoxide has a $-\frac{1}{2}$ charge per oxygen atom.  

The reaction is better thought of as two separate reactions:

\begin{align}
\ce{O2- &amp;&lt;=&gt; O2 + e-}\tag1\\
\ce{O2- + 2H+ + e- &amp;&lt;=&gt; H2O2}\tag2
\end{align}

This demarcation is particularly relevant in a biological context, because there is a class of enzymes Superoxide reductase (SOR) that only catalyze the reaction forming $\ce{H2O2}$ and another class of enzymes Superoxide dismutase (SOD) that catalyzes both. 

Here, oxygen is the element which is simultaneously undergoing oxidation (equation (1)) and reduction (equation (2)). This is termed a disproportionation reaction.

See also: A special case of disproportionation (or 'dismutation') is 'radical disproportionation'