Yes, a photon in a polarized light is found in a pure state such as $|H\rangle$, $|V\rangle$, $|L\rangle$, $|R\rangle$, or any complex linear combination of them. A photon in (completely) unpolarized light is described by the density matrix
$$ \rho = \frac{1}{2} \left( |L\rangle \langle L| + |R\rangle \langle R| \right) = \frac{1}{2} \left( |H\rangle \langle H| + |V\rangle \langle V| \right)$$
Note that you omitted the relationship for the vertically polarized state, $|V\rangle = i(|R\rangle - |L\rangle)/\sqrt{2}$, up to an overall sign which is a convention (well, the whole phase including $i$ is physically inconsequential, so it doesn't matter at all but one must be self-consistent with the conventions).
A magnetic dipole transition can be modelled as a time-dependent perturbation $V_{\text{md}}(t) = {e\over 2 m}(\vec{L} + 2\vec{S})\cdot \vec{B}e^{-i \omega t}$. Fermi's Golden Rule tells us that the transition rate for $b-X,1$ is proportional to the matrix element of the perturbation between the initial and final states,
$$W \propto \langle \psi_b|{e\over 2 m}(\vec{L} + 2\vec{S})\cdot \vec{B}|\psi_{X}\rangle,$$
where $|\psi_b\rangle$ is the excited state and $|\psi_{X}\rangle$ is the ground state (with three possible $M_S$ values.)
The effect of $\vec L$ and $\vec S$ will be to turn the final state into some combination of the triplet states, but it won't change $J$. Therefore we might expect the transition to be 'spin-forbidden':
$$W \propto \langle b^1\Sigma_g^+ |X^3\Sigma_{g,M_S=0,\pm1}^-\rangle = \langle J=0 | J=1 \rangle = 0.$$
This is where the spin-orbit coupling comes into play. Spin-orbit coupling is the reason why the singlet $(b)$ state has a higher energy than the triplet $(X)$ states. It is a perturbation of the form $V_\text{SO}={\mu\over\hbar}\vec{L}\cdot\vec{S}$, which can be rewritten as $V_\text{SO}={\mu \over 2\hbar}(J^2-L^2-S^2)$. In a spherically symmetric system like the helium atom, this perturbation commutes with the Hamiltonian, so all you get is a shift in the energy of the triplet (L=1) and singlet (L=0) states. However, in a linear molecule like $O_2$ you lose the spherical symmetry, so $[L^2,H]\neq0$ and in addition to an energy shift, you also get some mixing of the unperturbed eigenstates, so that the excited state is not exactly $|b^1\Sigma_g^+\rangle$, but rather $|\psi_b\rangle = c_1|b^1\Sigma_g^+\rangle + c_2|X^3\Sigma_{g,M_S=0}^-\rangle$. This mixing of J=0 and J=1 states is what allows $W$ to have a nonzero value. Since we can write $S_x = S_+ + S_-$, there will be a term in the transition rate like
$$W\propto c_2^*\langle X^3\Sigma_{g,M_S=0}^-|S_{\pm}|X^3\Sigma_{g,M_S=\mp1}^-\rangle+\cdots \neq 0.$$
Does this help? I know this is a bit hand-wavy so let me know if I can clarify anything.
Best Answer
Given two states $|n,l,m⟩$ and $|n',l',m'⟩$, the following radiative transition matrix elements are nonzero: $$\left\langle n,l\pm1,m+1 \middle| x+iy \middle| n,l,m \right\rangle,$$ $$\left\langle n,l\pm1,m-1 \middle| x-iy \middle| n,l,m \right\rangle,$$ and $$\left\langle n,l\pm1,m \middle| z\middle| n,l,m \right\rangle,$$ and that's it (for transitions between states of definite angular momentum about the $z$ axis; you can also have transitions to e.g. $|n,l+1,m+1⟩+|n,l+1,m-1⟩$, using a linearly polarized photon in the $x$ direction and propagating in the $y,z$ plane, or to similar superposition states).
The first two correspond to absorption or emission of a circularly polarized photon propagating along the $z$ axis, with selection rule $\Delta m=±1$ on the atom, while the third one corresponds to absorption or emission of a linearly polarized photon polarized along the $z$ axis and propagating on any axis in the $x,y$ plane, with selection rule $\Delta m=0$ on the atom.
For further details, consult any advanced undergraduate textbook on quantum mechanics.