How does an atomic transition between ground and excited states depend upon the direction of polarisation of incident light?
An electronic transition is characterized by its Transition dipole moment. To put it simple, it is a vector that shows the direction and magnitude of the electron cloud displacement. The probability of interaction with a photon is proportional to a scalar product of the transition dipole moment and the photon polarization. Electronic transitions with zero dipole moment are thus called forbidden.
Since atom is spherically symmetric, there is no given direction and the transition dipole moment points everywhere. Hence, there will be no preference to the polarization of light. However, you can impose a direction, for example, by applying electric field. Then the transition dipole moment will be oriented and the interaction with light will depend on its polarization.
You can't really talk about the relationship between excited atom states and incoming light polarization: the atomic states are what they are, independent of the incoming light. However, different transitions between atomic states can have probability amplitudes that depend on the incoming light's polarization: it's easy to see that the electrons in a molecular bond will interact more strongly with an electric field aligned with the bond.
If you're talking about incoming light provoking photon emission, well that's stimulated emission, and basic symmetry considerations show that the emitted light's polarisation must be aligned with that of the incoming light. See the Wikipedia page on stimulated emission and also the hyperphysics pages on the same topics and also on the Einstein A and B coefficients.
To think about what happens when light puts atoms/molecules into excited states and then the atoms/molecules spontaneously emit light sometime afterwards, one applies the principles of conservation of energy, momentum, and angular momentum to the light/atom/environment system as a whole. Polarisation relationships are about conservation of angular momentum. To understand this last statement, see, for example, the chapter called "Angular Momentum" in the third volume of "the Feynman lectures on physics". What does this mean practically? We think of three different cases: