Implication of negative order on kinetics
What does this $[\ce{O2}]^{-1}$ part mean?


It means the higher the concentration of oxygen, the lower the rate. The two-step mechanism you show relies on oxygen atoms (radicals) to react with ozone in the second step. Dioxygen acts as a radical scavenger (first reaction, reverse direction). So if dioxygen concentrations are high, the reaction will not proceed via this mechanism.

When ozone is produced as part of smog in the lower atmosphere (i.e. the air we breathe), this mechanism is not expected to help lower the ozone concentration. The equilibrium of the first reaction would be too far on the side of the reactants.

In the higher layers of the atmosphere, however, this sequence is more feasible. Concentrations of all species are extremely low, so bimolecular reactions are slow as well. In fact, in the image linked below, concentrations are given as molecule per mL (as apposed to mol / L) and rate a similarly expressed as molecules reacting per time unit.

https://www.ems.psu.edu/~brune/m532/meteo532_ch7_stratospheric_chemistry_files/image018.jpg

What actually happens in the ozone layer is a bit more complicated than the two step process you show, see for example https://www.ems.psu.edu/~brune/m532/meteo532_ch7_stratospheric_chemistry.htm


  For example, initially when there is no O2, the rate should be ∞; how is that possible? Or is it normal and I should accept it?


See the excellent comment by Nicolau Saker Neto. Another way to say this is "the molecules don't know the concentration of species, but they still do their thing without worrying about division by zero".