Shaded pole motor coil calculation, rewind 110v motor to 220v
Double number of turns would be right, but you want to half the area of the wire. or 1/sqrt(2) the diameter. 

New wire should be about twice as long. (More turns fit on a layer, and each layer is thinner so the next layer is smaller than the original). How close you get to this depends on how well you wind (and how well the original was!) 

Impedance will then be 4x the original, which is expected : at twice the voltage, for the same power, you draw half the current, so the power loss hasn't changed.

It looks as if there's plenty of room on the bobbin, so use the next wire guage up if you can't find an exact one, and that'll bring the resistances back down.

It's not clear exactly what you mean by "the active impedance of the wire" but in a running motor, only a small percentage of the AC supply actually causes current flow in the coil; the remainder is cancelled out by the "back EMF" from the rotor acting as a generator. (Until you stall the motor...)

Finally, if your 220V supply is 50Hz rather than 60Hz, expect the motor to draw more power than intended; it may not be running fast enough to generate the back-EMF it was designed for. I can't suggest how to adjust the turns ratio for that case.