Questions in MO theory
The antibonding orbital doesn't necessarily tell you the number of unpaired electrons - it's coincidental that for O$_2$, the HOMO (highest occupied molecular orbital) is a (1$\pi_g)^2 $ orbital - or, if you prefer to write it like you have for N$_2$, ($\pi ^1 _{2px} = \pi ^1 _{2py}$). 
$\pi$ orbitals are doubly degenerate, so the lowest energy configuration is for the 2 electrons to occupy different orbitals (one in each of the 2 $\pi$ orbitals). This means they're both unpaired, as shown in this MO diagram of O$_2$ from Wikipedia. For an example of an MO diagram where there are unpaired electrons in an orbital which is not antibonding, try drawing the B$_2$ MO diagram (remember that s-p mixing occurs).



In terms of what they do, antibonding electrons destabilise the bond, "cancelling out" the energetically favourable effect of bonding electrons - try drawing an MO diagram for He$_2$ or Ne$_2$. It should be apparent why they don't form stable molecules.

As far as N$_2$ goes, I think you may have neglected to take into account the effect of the $\sigma ^* _{2s} $ electrons. Remember that antibonding electrons cancel out bonding electrons.