Imagine for example the H$_2$ molecule as two spheres connected by a spring. Now imagine an off-center collision of two such molecules in a way so that only one of the spheres from each molecules end up participating in the collision. Since one of the spheres of each molecule sees a change in momentum while the other does not, they will start diverging. However, their coupling will try to pull the diverging molecules back together. Depending on the details of the collision, the two-sphere molecule will generally start rotating about its center of mass as well as oscillating. This is the transfer of translational energy into vibrational and rotational degrees of freedom.
Once you start considering collisions where the molecules are already initially vibrating and rotating, you will realize that collisions can also transfer rotations and vibrations into translational energy and also rotations and vibrations between the molecules. There is really no restriction between the channels, every form of kinetic energy can be transferred to any other one.
As for the activation of the vibrational degrees of freedom, this requires to think about quantum mechanics. This is because if your molecules were purely classical, then no matter how stiff the ``molecular spring'' would be, vibrational degrees would be activated (and would receive their fair share of $k_\mathrm{B} T/2$ of the energy on average).
The point is, however, that the spacing of the quantum energy levels is $\hbar \omega$, where $\omega$ is the vibrational frequency of the spring. We can then estimate the activation by considering the partition between the ground state of the vibrations and the first excitation. The relative proportion of the first excitation will then be $$\approx \exp\left(-\frac{\hbar \omega}{k_\mathrm{B} T}\right)$$
at least at low temperatures. We see that whenever $\hbar \omega \gg k_\mathrm{B} T$ the proportion of molecules with excited vibrational degrees of freedom is negligible. On the other hand, whenever $k_\mathrm{B} T \gg \hbar \omega$, the occupation of all the vibrational levels will be high and we can even go to the classical limit of the vibrational degree of freedom. (Then again, when $k_\mathrm{B} T \gg \hbar \omega$ the harmonic approximation of the spring will likely be breaking and you might even be dissociating your molecules at appreciable rates).
As for radiation, that is not really different. Collisions as well as relaxation of the vibrational degree of freedom will generally produce photons. If the ensemble is in thermal equilibrium, you will have a photon gas inside it taking up a fair share of internal energy. Depending on the conditions, this may or may not form an appreciable fraction of the pressure and or internal energy of the system.
The only thing that can be relevant I can think of in this context is the fact that the mean life-time of the first vibrational excitation is $\tau = 1/\omega$. If this time is considerably shorter than the typical time between collisions, then the vibrational excitation can be assumed to be out of thermal equilibrium and damped by the radiation (at least if radiation is leaking from the system). However, this will usually occur only for gases where the fraction of thermally excited vibrations would be negligible already by the previous argument.
