In non-mathematical, hand-waving terms, orbitals are mathematical representations of states in which electrons can exist. A single atom, with one nucleus and a multiplicity of electrons, will have the electrons in orbitals which provide the lowest energy for the entire system. You are familiar with those types of orbitals, obviously.
When you add a second nucleus at an appropriate distance, you can imagine a new set of orbitals which are vector sums of the orbitals on a single nucleus system. As Juanrga said, they are the a + b and a - b sums. One set will be slightly lower in energy than any of the orbitals they were summed from (stabilizing, bonding orbitals), and the other will be slightly higher (de-stabilizing, anti-bonding). In general, the antibonding orbitals are somewhat more destabilizing than the bonding ones are stabilizing (due to entropy).
The orbitals are occupied in the usual way, from lowest to highest energy. This means the electrons will be occupying the bonding orbitals first, then the anti-bonding. If there are more electrons in bonding orbitals than in anti-bonding orbitals, the result will be that the two-nucleus system is more stable than the separate single-nucleus systems, and the nuclei will be bonded. The most stable bond would be when all of the bonding orbitals are filled and all of the anti-bonding orbitals are empty - that would be the highest level of stabilization possible from the combination. However, if there are as many electrons in anti-bonding orbitals as there are in binding orbitals, then the slight excess of destabilization energy means that the net energy of the system is higher than it would be if the two nuclei were separate, the bond is not stabilized, and the nuclei go their separate ways.