Molecular orbitals of CH2 an SLACs
The first question to always ask when you are trying to build an MO diagram is what point group does my molecule fall into?  In this case, the $\ce{CH2}$ molecule has C2V symmetry (like the water example shown above).  Each of the atomic orbitals (AO's) on carbon (left) and the combination of $1s$ orbitals on the hydrogen atoms (right) has a symmetry label that corresponds to one of the irreducible representations of the point group.  

You can determine by symmetry label for the AO's by operating on the orbital with each of the symmetry operators in the C2V group (E, C2, σ, σ').  For example, the 2py orbital is in the plane of the nuclei and so it is symmetric with respect to E and σ' but anti-symmetric with respect to C2 and σ and has b2 symmetry from the C2V character table.  

For the hydrogens, you have to consider the combination of both 1s orbitals at the same time.  The a1 combination is when the phase of the wavefunction is the same and the b2 combination is when the phase is opposite (i.e. H1 is positive and H2 is negative or vice versa).

AO's that have the same symmetry can combine to form MO's with that symmetry label.  In the above example, the 2s and 2pz orbitals on oxygen have a1 symmetry, and they combine with the a1 combination of hydrogen 1s orbitals to form three MO's with a1 symmetry.  The 2py (b2 symmetry) orbital combines with the b2 combination of the hydrogen atoms to form two MO's. Finally, the 2px orbital remains unchanged because there are no other SLAC's with b1 symmetry.  

The MO's are qualitatively ordered based on the number of nodes that are found when you combine the SLAC's.  The numbers in front of the MO's differentiate the orbitals with the same symmetry.