Does photon absorption occur in discrete packets? As per the dogma of quantum mech, the jump of an electron from a higher to a lower energy shell is accompanied by photon emission. All photons are born this way. (Although an off topic rebuttal might say that there is no reason the same process cannot occur with protons in the nucleus and might account for the nuclear flash of those explosions). Anyway my question has to do with the ‘death’ of photons. Certainly absorption of photons by atoms can result in several different manifestations of increased atomic energy. Vibration, rotation, linear motion(increased temp), and even excitation of the electrons to higher energy levels with spontaneous re-emission of new photons at different frequencies are possible. So like us, photons can die in many ways.
But is there any data that looks at the energy of an atom when a photon hits it?
Is the energy jump a quantum (discrete and reproducible amount whose value is part of a limited set) amount?
I thought about the impossibility of measuring this. Can one measure the energy of an object unless it does something? And even though the atom might ‘swallow’ photons in discrete packets, the manifestation of that energy is not necessarily quantal (vibration, rotation, etc)
But it seems almost BY DEFINITION absorption must be quantized because you can never measure a fractional (leftover) photon. But that is only if we consider absorption by single atoms. However a molecule of multiple atoms could also absorb a single photon. Then which atom of the molecule gets its electrons excited? Since the electrons of a covalent bond are shared, excitation of those electrons might be considered’fractional’ absorption. But this is just semantics- it is the molecule that does the absorbing so does it even make sense to ask how the energy is distributed among the atoms?
Fractional absorption has another possibility of manifesting. If the atom is traveling at relativistic velocity then its absorption of energy is limited by the speed of light. So even as the relativistic atom is bombarded by more photons, shouldn’t its energy jumps (quantized though they may be) become smaller at higher velocities?
 A: 
Then which atom of the molecule gets its electrons excited?

This question stems from an incorrect understanding of how molecules work. A single molecule is an object where all the constituents: electrons, nuclei, — are entangled with each other. There's no single-electron or single-nucleus nor a single-atom wavefunction that could describe a part of a molecule. Whenever we analyze emission or absorption of EM radiation by a molecule, it happens as a transition from one eigenstate of the whole molecule to another. A good example to illustrate this is photoisomerization, which is the principal photon-triggered reaction driving visual phototransduction in animals.

does it even make sense to ask how the energy is distributed among the atoms?

If you can devise such an observable, then yes, it does make sense. In this case, each electron and each nucleus has kinetic and potential energy operators, so you can use these to answer your question.

If the atom is traveling at relativistic velocity then its absorption of energy is limited by the speed of light.

Well, not quite. Yes, it's affected by time dilation & length contraction, in the sense that energy levels change, but this doesn't "limit" energy absorption. It just results in sensitivity to different photon frequencies than a resting atom has.
But, as another answer mentions, a photon can be "partially absorbed": this is called inelastic scattering, where an incident photon is scattered getting a different final energy, giving a part of its energy and momentum to the scatterer. Examples are Raman scattering and Compton scattering.
A: Yes. If a system's energy levels are discrete, then it's absorption spectrum is also quantized.
Absorbing half a photon is not possible.
Many analytical techniques leverage this effect.
Also, yes, if the atom is moving with respect to a stationary observer, then it seems as if the energy levels are shifted. This is simply a Doppler shift, and it is exploited e.g. in astronomy to measure distances and velocities of far away objects.
The only thing where I have to correct you is that not all photons are created from the relaxation of an excited electron in an atom. There's plenty of other processes available, including antennas or the nuclear relaxations that you mentioned.
A: Photons are indivisible quanta off energy from their emission to their absorption. In any case a re-emission takes place after the absorption.
According the second law of thermodynamics the re-emission always is with photons of lower energy plus low energy photons (for our surroundings in the IR range).
A: 
But is there any data that looks at the energy of an atom when a photon hits it? Is the energy jump a quantum (discrete and reproducible amount whose value is part of a limited set) amount?

The reason quantum mechanics became necessary was because classical mechanics and classical electrodynamics could not explain the black body radiation, the photoelectric effect and, the spectra of atoms. These last are the data you are seeking, here in the visible electromagnetic wave spectrum.


Continuous, emission, and absorption spectra

They were explained with the Bohr model, a precursor of the full theory of quantum mechanics.
In general, the photons  are quanta of energy that make up in a quantum mechanical superposition the visible, in the above case, light. Photons are elementary particles that have energy h*nu, where nu is the frequency of the classical light they build up. In general they can be produced  in various ways, not only in atomic transitions, and give a continuous spectrum.
Photons can scatter elastically or inelastically with atoms and charged particles, and these interactions have been  studied .
