Lost track of the hydrogen in step 3 of the Krebs cycle
Your problem stems from the statement
&quot;I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+.&quot;
This statement is incorrect. In isocitrate, the carboxylic acid functional groups are deprotonated, as would be the case at neutral pH. The oxidation reaction thus involves loss of only $\ce{H-}$ to NAD. There is no loss of $\ce{H+}$, because that has already been removed.
If you start with isocitric acid, which is isocitrate in its neutral form with $\ce{-CO2H}$ instead of $\ce{-CO2-}$, then you will need to remove $\ce{H-}$ and $\ce{H+}$ to accomplish the reaction and your above statement will be correct.
The textbook image you showed takes a confusing approach of implicitly associating an $\ce{H+}$ with isocitrate even though it is not actually on the molecule, so that &quot;NADH/H+&quot; is indicated as a product even though only NADH has actually been produced in the reaction shown.