Four stroking is when a 2 cycle is gagging on too much fuel. As soon as you apply a little bit of load it needs more fuel, hence cleaning up in the cut. Amount of work being done determines how much added fuel is needed. That's kind of how I look at it
Gagging? More work = more fuel? Simple terms are good, but when I argue a point I try to include all necessary details to prove the general concept. That's where repetition of simple terms in an argument bugs me, because it's like the same exact explanation is supposed to change my opinion when no new supporting facts are presented. So, then I try to induce a more detailed response by representing my arguement in further detail to either convince, or to provide the precise details to be countered to convince me of something different.
Back to my 4 stroking theory, but first my response to More work = more fuel
Gas engines are not like electrics. They don't efficiently use as much fuel as is needed, they use as much fuel as you give them: whatever the carb needles are set to, and whatever volume of air the current throttle position allows, of course according to rpm as well.
Electric motors are kind of self governing in a way. Open the throttle circuit to max (full bore of solid wire / lowest resistance or 100% on duty cycle) and at 0 rpm the motor is drawing max power to push the magnets. The faster the rotor spins, the less it load it sees like a torque converter, until max rpm is reached. If there is no load at max rpm, other than various frictions, the load is minimal on the motor, it doesn't have to push the magnets as hard because they are nearly matching the speed vector of the electro magnet (I really wish I knew enough about them to explain more precisely), and so minimal amperage (fuel volume) is drawn. Spin the motor as fast as it wants to go, and it will draw no power at full throttle. Spin the motor faster than physics says it should go, and it will output power back into the battery.
Run a gas engine without load, and it pulls the same volume of fuel per revolution (actually rpm dependent based on several variables, but the example is plenty true enough). How much power is it putting out? Well, the gas is releasing most of it's potential energy depending on how clean it burns, *but considering the limited rate at which gas burns, and engine rpm, and richening of the mixture due to high rpm*, the power output at the shaft is lower than it would be under load.
Run a gas engine under load, and it does not draw any more fuel. Technically it should draw less volume over time due to a lower rpm than no load revving, but still roughly the same volume per stroke, a little on the leaner side now that the high rpm isn't demanding too much CFM through the carb.
As for more work = more fuel, I agree 2 stroking during a no load rev is too lean for loaded use, so use more fuel for more work, but as for 4 stroking cleaning up to 2 stroking in the cut, I have some pictures to better represent my theory.
Imagine the fuel air mixture burning at the exact same speed no matter the rpm.
You can look at the picture in reverse for the effect of 4 stroking cleaning up into 2 stroking as load increases and rpm decreases. It all has to do with the burn speed of the fuel air mix relative to the time that the transfers are shut base on rpm.
* * = more detail can be given if necessary.
. no really 
,,, thinking about it 4strokin is the mix is rich, we know this cos revs climb if lean.... so it’s a ratio of fuel and air... more fuel than optimal, but not lean... stick saw in a cut and fuel has a bit more time to burn and cleans up...
