Ok, my guess was wrong, but that solution does use the momentum argument.
The author considers an element mass dm going from stationary to moving at speed v0 in time dt.
Since the upper rope is moving at speed v0, in time dt it moves to the right v0dt. But that only advances its left hand end by v0dt/2, so dm= λv0dt/2. That's where the v0/2 comes from.
Since that dm went from 0 to v0, its momentum increase was λv02/2 dt. Hence Fdt=λv02/2 dt, so F=λv02/2.
However, if we assume work is conserved we get F=λv02/4. I'll leave it to you to see how.
So which is right?
It's always risky assuming work is conserved, and one can see ways in which losses occur in this set-up. But to lose half the work?
It turns out the momentum argument is flawed. It assumes the length of rope on the table exerts no force on the rising element; but that is surely false. The rising segment has to rotate about its lower end.
If you push one end of a stick lying on a smooth surface at right angles to the stick its centre of rotation is not the other end of the stick, but a point somewhere between there and the middle of the stick. For it to rotate about the end, you need a bit of a push on that end too.
We can deduce that the resting section of rope must exert some tension on the rising element, thereby assisting F.
On balance, I would say option (4) is closer to the truth than option (3).