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.
The selection rule $\Delta S=0$ is an approximation, nothing more, and in suitable circumstances it can easily break. One prominent example of this is the hydrogen 21cm line.
Electromagnetic atomic and molecular transitions are arranged into a series in order of multipolarity, which describes the atomic operators that enact the interaction hamiltonian, with higher multipolarities scaling down in coupling strength as powers of $a/\lambda$, i.e. the ratio between the system's size and the radiation's wavelength, which is generally small. Thus, you get
- as the leading-order term, electric dipole (E1) transitions;
- weaker by a factor of ${\sim}a/\lambda$,
- magnetic dipole (M1), and
- electric quadrupole (E2) transitions;
- still weaker by another factor of ${\sim}a/\lambda$,
- magnetic quadrupole (M2), and
- electric octupole (E3) transitions;
- and so on.
Generally, the no-spin-flips rule holds only for electric dipole transitions, for which the interaction hamiltonian is the electric dipole operator $\hat{\mathbf d}$, which does not couple sectors with different spin (unless you have strong spin-orbit coupling).
However, it is perfectly possible to have magnetic dipole transitions between states of spin that differ by $\Delta S=1$. Here the coupling is weaker (so you will need a higher intensity, or longer pulse times, to excite them), and thus typically the linewidth is smaller (so you will require a sharper laser), also giving a longer decay lifetime, but those are things that make the transition harder to observe, not impossible.
Best Answer
The spin of the electron has to do with the deflection in magnetic fields and with the orientation between electrons. The spin of the photon has to do with the direction between the photons electric to its magnetic field component. In this sense these spins are totally different things.
Only a rotating electron passes on a part of its torque to the emitted photon. The two E and M field components of the photon are then (in vacuum) still oriented 90° to each other, but rotate together in the axis of motion of the photon.
Polarization of photons occurs when charges are accelerated in the same direction. If the photons are accelerated upwards, then the electric field component of the emitted photons also points upwards and the magnetic field component is perpendicular to E (and to the propagation direction of the photons). This is the process as it is actually detectable for antenna radiation.
What I don't know - but is worth a question on PSE - is whether the spin of the electrons aligns during accelerations (in the antenna rod). If so, then it is prediction that the spin tilts towards the perpendicular with respect to the acceleration direction.