Treat the non-linearity+noise as a composite noise signal, \$v_n\$, then the output will be: \$v_o=v'+v_n\$, where \$v'\$ is the op-amp output signal. Also assume, for simplicity, that the potentiometer is in the middle of its travel.
The signal fed back to the inverting input, and consequently equal to the input voltage, is then: \$v_i=v_o-\frac{v_n}{2}\$, giving: $$v_o=v_i+\frac{v_n}{2}$$ Thus the output noise is halved.
If the feedback is entirely from the \$v_o\$ end of the potentiometer, the noise is removed completely; if it's from the op-amp output, all the noise appears at \$v_o\$