I am not sure if this will make sense but here goes:
In PbO2, the oxidation state of the lead is +4, because the overall oxidation state of PbO2 is 0 because charges cancel out, leaving a compound neutral. The oxidation state of O is always -2, and there are two oxygen atoms here, so the charge on the O2 is -4. Therefore to make PbO2 0, the oxidation number of Pb must be +4.
If something has a charge of +4, this means that it has lost 4 electrons, right? (think about sodium as an ion - Na+) So surely in order to become Pb2+, it has to have lost 2 electrons. This means that it has to have gained 2 electrons in order to become Pb2+ on the other side of the equation.
And it's the PbO2 that has to have gained 2 electrons in order to become Pb2+, because if you think about it, if the + 2e- was on the other side of the equation, that would be equivalent to the PbO2 losing two electrons, which would not make Pb2+, but Pb6+, which isn't what you want as a product.
Hope that helps - please reply if that explanation was ok
