You asked three questions in one sentence, and then more in the following, I will attempt to address each individually
Assume that you have one electron in the excited state, you have only one electron spinning in only one direction therefore
(1/2)*2 + 1 = 2 ,
you have something in a doublet state.
Now, let us assume that two electrons went from the ground state to the excite state (S0 to S1).
The difference between a triplet and a singlet state is the direction of spin of the TWO excited electrons.
For a singlet state, spins are in opposite directions , therefore we have one electron moving in a positive direction and the other moving in a negative direction making n, (which I will use to represent spin of the electron), making n equal to zero.
2n + 1
2 * ( +1/2 -1/2) + 1
2 * 0 + 1 = 1 (singlet)
For a triplet state, spins are in the same direction; making the spins of the electron (1/2 + 1/2) making n = 1
2(1) + 1 = 3 (triplet)
Transitions in a Jablonski Diagrams may occur from the ground Singlet state (S0) to the Excited Singlet state (S1) Transitions from the ground singlet state to the triplet state are rare, and are called forbidden transitions typically an electron is excited to S1 then releases heat and goes from the Excited Singlet state to the Excited Triplet State.
When something goes from excited singlet state to Ground state, energy is released and called Fluorescence, from the excited triplet state to Ground state is called phosphorescence.
Phosphorescence tends to dominate in the presence of strong oxidizing agents ( for example H2O2)
This is the theory, now to tie it all back to luminol you will need some one with a PHD, but I'm guessing that after the Nitrogen is produced, that the electrons on Oxygen are in pi* before transitioning to pi,
I am a student myself and am confused as to why luminol will fluoresce for so long, it would make more sense to me if it phosphoresced, perhaps it is simply a convenience of vocabulary that keeps luminol fluorescing.