[eddyfm11]

For example, in the following CO₂^2- ion, in each of the resonance forms, there is one oxygen atom which have 2 lone pairs, and two oxygen atoms with 3 lone pairs.

When they form the resonance hybrid, however, the molecule is depicted such that each oxygen atom only has two lone pairs of electrons.

It is clear to me from what I've read that the pi bond electrons are delocalized (shown by the dotted lines). However, are the lone pairs assumed to be delocalized? If not, where are they in the resonance hybrid? (Only 10 electron pairs are depicted in the resonance hybrid, with 2 lone pairs missing.)

[mjc123]

The short answer is you can't draw fractional lone pairs in a Lewis structure, so it's impossible to represent them accurately. As you can see form your resonance structures, each O has 2 2/3 lone pairs - if that is at all meaningful. So yes, some of the lone pairs participate in delocalisation.
Draw your basic σ bond structure (sorry, I can't do this in Paint). You have three Os each with 3 lone pairs and a - charge; the C has a + charge. Two lone pairs on each O are in sp² orbitals, the third in a p_z orbital. The p_z on C is empty. Four p orbitals combine to give 4 molecular orbitals - one bonding, two non-bonding and one antibonding. Six electrons fill the bonding and non-bonding orbitals, so you have effectively one pi bond, or an average order of 1/3 per C-O.