Bode plot: meaning of phase shift
Is it the shift of the output compared to the input signal? 


It depends what you're making a Bode plot of.

If you're plotting a transfer function \$V_{out}/V_{in}\$, then yes, it will be the shift of the output compared to the input.

If you're plotting an impedance \$V_{in}/I_{in}\$ then it's the phase shift of the voltage compared to the current.


  I don't really understand why the shift goes to zero as the angular velocity (omega) goes to infinity


Because in the limit, the capacitor becomes a short circuit and the inductor becomes an open. So the equivalent circuit is just connecting the input directly across the load resistance.


  and why the amplification goes to zero as omega goes to infinity


Same reason. The equivalent circuit in the high frequency limit is just applying the input directly across the output. 0 dB gain is equivalent to linear gain of 1.