Disproportionation of 2 KO₂ + 2 H₂O → 2 KOH + H₂O₂ + O₂
The disproportionation happens to elements, not to compounds.

In this particular case, to the part of oxygen,

$$  \overset{\ce{2 KO2}}{ 4 \times -0.5}=  \overset{\ce{H2O2}} {2\times -1} +\overset{\ce{O2}}{ 2\times 0}$$

while the other part keeps the oxidation state constant:

$$  \overset{\ce{2 H2O}} {2 \times -2}=  \overset{\ce{2 KOH}} {2\times -2}$$

$\ce{H2O}$ does not get directly involved in disproportionation. 

$\ce{KO2}$ oxidizes and at the same time reduces  itself. 

This is the nature of isproportionation, which happens , if the particular compound of the element in the intermediate oxidation state has higher Gibbs energy then particular compounds of this element in higher and lower oxidation states. Therefore redistribution of oxidation states is thermodynamically preferred.

Typical cases are $\ce{2 Cu^{I}-&gt;Cu^{II} + Cu^{0}}$ and $\ce{2 Hg^{I}-&gt;Hg^{II} + Hg^{0}}$