In response to your first question, the water doesn’t add negative angular momentum because the whole system we are conserving is the sprinkler body and the moving water together. The sum must just be zero in both cases. We can track a single drop of water in case two (Feynman) to see how the conservation works. At the start the drop is at zero angular momentum since it is stationary in the tank. It then passes through the whole Sprinkler mechanism (and whatever s-curves there are) and ends up at the suction drain at the center of the sprinkler also at zero angular momentum. Therefore no matter what happened it could not have applied a net torque to the sprinkler, no matter the path it took to get to the drain.
Contrast with case 1, the regular sprinkler, where the water drop starts at zero angular momentum (pumped from the center of the sprinkler) and ends with positive angular momentum (spinning away from the center) therefore it must have applied an opposing net torque at some point in its journey to the body of the sprinkler, which spins.