The contribution of the kite to lifting the rider AFTER the launch is entirely missed in this model. It is this part of the equation that IS dependant on the mass of the rider
i'll explain again. This particular model does not depend on the mass of the kiter.
there are two vectors.
1) kiter in relation to water
2) kiter in relation to wind
1) For the kiter in relation to the water, he pops. His vector is directed upward. It does not depend on his mass any more than a BMX biker doing a jump depends on his mass.
2) At the same time the kiter redirects the kite to boost.
The model in this case says that the kite has the same mechanical properties as a tack stuck in the wall with a lead weight swung from it. If the tack is big enough (if the kite is big enough), then the lead weight's direction will change from one direction to the other. It does not matter how big your lead weight is, unless you decide to put in a tiny little tack (under power the kite).
Think of it as sticking your kite in the air. You have already popped out of the water, and all of your horizontal motion relative to the water has been converted upwards. But the kite (and yourself) still have a lot of speed relative to the speed of the air, and you want to convert this to vertical height too. You use your kite in the air as a pivot to change your horizontal motion into potential energy. Then both you and the kite rise up in the air.
Now I'm not saying this model is perfect. But it does say, if you have enough kite, it doesn't matter how much more kite you won't boost higher.
So for instance suppose you go out and it's 25 knots and you are lit on your 10 meter.
Then you switch to your 20 meter, and you get worked up and down. You go to jump anyway.
Do you go twice as high on your 20 meter as you did your 10 meter?
Or did the 20 meter just let you drift down a lot slower?
Anyway, I'm not defending this model, just trying to explain it properly so you understand why your objections are not proper objections, they stem from a misunderstanding of this particular model. Even though vectors 1 and 2 come from different places you can just add them together for 1/2mv^2 = gmh, where v is velocity relative to wind, that's how vectors work. Notice the 'm's the mass cancels out here.