What is the "leakage current" between signal and shield?
1) When the shield isn't connected to anything, how can any current be there and where will the current flow to?

The shield is ordinarily connected to ground or some other low-impedance path to ground. If it wasn't, it wouldn't be that useful in stopping noise. With such a connection, the (grounded) shield forms a capacitor between itself and the leads inside, biased at the common-mode voltage.

Because no insulator is perfect, there is a small leakage current as well, modeled as a resistance to ground. The greater the voltage difference between shield and leads, the greater the leakage.

2) How does connecting the shield to common mode voltage reduce both leakage current and capacitance?

Driving the shield to the common-mode voltage reduces the DC bias between the shield and the signal lines. With a smaller (or practically no) bias, there is no DC leakage. It also cancels out the common-mode AC voltage so it doesn't influence the signal pair either.

Some relevant info:


biometrics: https://www.biosemi.com/publications/pdf/Interference_reduction.pdf
instrumentation: https://www.analog.com/media/en/technical-documentation/application-notes/41727248AN_347.pdf