Let's think about how the two polarization components act on the atoms independently first. I am assuming here the propagation direction of the beams is parallel to the quantization axis of the atoms.
The two frequency components have polarization $|V\rangle$ and $|H\rangle$, respectively. Linear polarization propagating along the quantization axis of the atoms will drive $\sigma^+$ and $\sigma^-$ transitions, so we should decompose the linear polarizations into circular: up to normalization and sign conventions, we can say $|V\rangle = |R\rangle + |L\rangle$ and $|H\rangle = -i(|R\rangle - |L\rangle)$ (where $|R\rangle$ and $|L\rangle$ are right handed and left handed). The $|R\rangle$ components drive $\sigma^+$ transitions and the $|L\rangle$ components drive $\sigma^-$ transitions.
Now, suppose both components have the same frequency but one component has phase $e^{i\pi/2} = i$. The first component $|V\rangle$ drives $\sigma^+$ and $\sigma^-$. The second component $i |H\rangle = |R\rangle - |L\rangle$ drives $\sigma^+$ and $-\sigma^-$ transitions. The $\sigma^+$ driving constructively interferes and the $\sigma^-$ driving cancels out - we are left with just $\sigma^+$ driving as if the combined field were purely circularly polarized (which it is!)
But this analysis now lets us extend to the case where the two components have different frequencies. If the frequency separation is small compared to the driving Rabi frequency, then as far as the atoms are concerned there is no difference compared to if they have the same frequency. But if the frequency difference is comparable to the other energy scales in the problem, then you should consider what transitions each beam independently drives and then add up all the driven transitions (including their phases!) to see what cancels out and what ends up being important.
