First, it would be much better if you post the equations you're using in plain text, as it's very difficult to read your hand-writing (my eyes are old, anyway), so I'm not at all sure I'm correctly interpreting your question or the equations show in the picture.

My guess: In the diagram just below "calculate bounce from ground," it appears you have angular velocities Wx1 and Wx2 defined as positive** when pointing along the positive X direction (left-hand rule). However, your torque equation R x F is right-handed - i.e., R X Fz (left-handed) should result in a +X torque, not a negative X component, if R==(0, -r, 0) and Fz==(0, 0, -Fz).

**assumed because you show a CW rotation about the X axis.

E.g., ( 0, -1, 0 ) cross ( 0, 0, -1 ) = ( 1, 0, 0 ) [left-hand rule]

You must define torque, angular acceleration, and angular velocity all with the same handedness.

EDIT: If you want to use a right-hand rule for torque, then use F x R.

Yes, Wx1 and Wx2 is > 0.

I understood my bug. The bug was that I calculated RxF like (0,-r,0)x(Fx,0,Fz) instead (0,-r,0)x(-Fx,0,-Fz).

Thanks for help.