How exactly does the oxidation of TEMPO by NaOCl/KBr work? I would like to understand the exact process by which TEMPO is oxidized to the nitrosonium cation by NaOCl, as in the part of the mechanism shown in the diagram (1) surrounded by the red box:



The confusion for me is that TEMPO clearly needs to donate 1 electron to $\ce{OCl-}$ in order to form the neutral, nonradical nitrosonium cation, but we also get a $\ce{Cl-}$ anion, so there must also be a 2 electron process involving the $\ce{Br-}$ "co-catalyst" coming into play. How do I reconcile these 1 electron and 2 electron processes so as to draw a formally appropriate mechanism for this part of the reaction?

What exactly happens to the $\ce{OCl-}$ oxygen, and how does it happen?

(1) Org. Synth. 2005, 81, 195. DOI:10.15227/orgsyn.081.0195.
I would propose the following mechanism:

Note that the oxygen radical anion isn't that bad as a leaving group as the Cl-O bond is weak. It can participate in an electron transfer with chloride anion, or rather the bromide at the beginning of the reaction as it is more abundant (I have written the protonated form - hydroxyl radical as it is less likely that two anions will come close enough for electron transfer):


The bromine radical can in turn react with another TEMPO molecule. That is why two equivalents of TEMPO react with one equivalent of hypochlorite.

Keep in mind that there are countless possibilities and deviations from the mechanism I proposed.