A photon is emitted by an electron (which is in a bound state). Is the energy of the electron lost immediately, or is the energy emitted during the complete transition time? I think my second assumption is correct but confirmation would be greatly appreciated
One has to let go of the classical framework. Immediately in time has a meaning for a ball falling in the Newtonian gravitational field and mathematics can give you the rate of energy loss per delta(t) because in principle every (x,y,z) point is reached at a specific t. This does not hold in the quantum mechanical framework of an electron bound to an atom.
An individual atom with its electron in an excited state may emit the photon at an arbitrary time t. One has to take a large sample of atoms with the electrons at that energy level and measure the time the photon hits the detector . One then will have a curve characterizing the lifetime of that bound state's collective time behavior
Now in the post you have quoted the answers are indicating the mathematical formulation within the theory of Quantum Mechanics that reproduce this experimental observation. Quantum mechanical calculations give probability distributions for the variables under consideration, time in this instance, to fit the experimental observations.
What is really happening at the individual atom's decay from a higher energy level to a lower one is random to first order ( it is the assumption of calculating the half life curves) . The time will be within the Heisenberg Uncertainty of delta(E)*delta(t) is all that can be safely claimed , and can be presumed "instantaneous" , in the sense that there are no experimental tools to explore further, other than the mathematical ones discussed in the answers of the other question.
