That's a pretty uniformed comment. If a film of water gas isolates thermally and presumably electrically the cathode then less current flows and less heating of the electrolyte occurs. If there is only a single spot of the cathode left conducting then the bulk electrolyte won't reach the temperature where it would readily boil. If you had any experience of electrolysers and plating tanks you would know that,
The phenomenon is progressive. For a long time the vapor bubbles don't completely isolate the cathode, so that current is allowed to reach the palladium, but contemporarily heat is partially hindered to leave it.
In any case the electric energy dissipated in Joule heating is known, because it is shown by the diagrams of the experiments: Figs.6A-D in http://www.lenr-canr.org/acrobat/Fleischmancalorimetra.pdf
for the F&P experiment, and Figs.3a-b in http://www.lenr-canr.org/acrobat/LonchamptGreproducti.pdf for Lonchampt's.
All these diagrams show an extraordinary increase in the absorbed power or voltage, which are proportional each other because current is kept constant. As the electrolyte temperature approaches the boiling point, the absorbed voltage(or power) becomes ten times greater than the initial values. It means that up to 90% of the final electric power can be dissipated as Joule effect in the cathode. This only fact, documented in F&P and Lonchampt diagrams, is sufficient to explain the overheating of cathodes.