Spin-forbidden transition iof bromine
The selection rules are related to the transition dipole moment that determines the intensity of a given transition. If a given selection rule holds perfectly, all that it says is that a forbidden transition has zero intensity and an allowed transition has non-zero intensity.

If the selection rule does not hold exactly, just like the in the case of the spin selection rule for heavy atoms, a forbidden transition can be intense. Some examples: the $\text{B}^3\Pi_u-\text{X}^1\Sigma^+_g$ electronic transition in $\ce{Br2}$ and $\ce{I2}$, that are strong when compared to the same transition in $\ce{F2}$ and $\ce{Cl2}$; and the strong $^3P_1-^1S_0$ transition in $\ce{Hg}$, while the analogous one in $\ce{He}$ is very weak. As $L$ and $S$ are not good quantum numbers when there is strong spin-orbit coupling, the selection rules involving them can't be used.

The second answer you came up with makes sense, but not for spin selection rule. Consider the parity selection rule for molecules with an inversion center. As the molecule vibrates (in an asymmetric mode) it can be distorted into a geometry that does not have an inversion center, so the selection rule does not hold. In this case, this vibronic transition will be allowed.