If an electron absorbs a photon to get exited to a higher energy level, it should either come back to same state or any other lower state by emitting the required photon. How then can there be a net transfer of energy to the atom? Heating up means increase in kinetic (vibrational) energy of the atoms. If the energy absorbed by a photon is re-emitted as a photon then how do atoms extract energy from incident photons?
Let us clear some misunderstandings:
It is not the electron that absorbs the photon to go to a higher energy level. It is the whole atom, which is represented by a potential well with energy levels filled by electrons up to a point. A photon with the correct energy, i.e. an energy that covers the difference of the level where the electron is and a higher energy empty level, will be absorbed by the whole atom. The electron will decay with a characteristic decay time from the higher energy level to a lower one and a photon will take up the energy again. It can happen that cascades of photons may take up the energy.
Again, it is the system nucleus+electrons that absorbs and emits quantized photons.
In this case, ( of absorption and emission), little kinetic energy is transferred to the atom, from momentm conservation within the Heisenberg Uncertainty Principle.
I hope it is clear that it does not. It could happen that, for example, the dipole field of an atom interacts with the field of another atom and transfers kinetic energy it has received, but it will be from continuum scatterings, not from quantized absorption and emission.
No, photons from the continuum can interact with the dipole and quadrupole moments of the atoms and transfer energy there. Infrared photons of the continuum can scatter off the left over fields of the atoms/molecules transferring momentum /kinetic energy and increasing temperature. It is a continuum process, not a quantized one.
right but also of kinetic energy
They do not, they gain it from the continuum of infrared photons interactions with the left over fields of the atoms and molecules.
Essentially what you are asking about is the "photoelectric effect". The intensity of the light does not contribute to energy needed for the material to expel a photon.
The Minimum energy needed to release a photon is given by the work function $\Phi=hf$ where h is Planck's constant and f is frequency.
Here is a flash simulation of what is going on.
What the affect the intensity has is that it will change the number of electrons that are excited by the incident light which is different.