Higher KV means lower torque for DC motors?
That site is misleading, a motor wound for low rpm/V (also known as KV) doesn't have greater or lesser torque handling than the same motor wound for high rpm/V. The real difference is that the low rpm/V motor requires more voltage at proportionally less current (same power) to spin a given load at a given speed. The efficiency, maximum torque and maximum power are still the same for both motors.
Replacing a high rpm/V motor with an otherwise identical low rpm/V motor only increases torque if the system is bottlenecked by the maximum current that the motor driver can deliver to the high rpm/V motor.

Thinner windings in motor windings do mean higher resistance, but this balances out with the lower current draw. For example, think of a motor wound with 8 turns of 2 mm^2 wire for each stator pole. The same motor, but wound to half of the rpm/V value, would have 16 turns of 1mm^2 wire instead. Yes, the wire area is cut in half, so you get twice the resistance per turn, but you also have more turns, with the wire being twice as long (16 turns vs 8). Thus, the 16 turn motor would have half the rpm/V and 4 times the resistance.
This is not as bad as it seems. The 8 turn motor will have to draw twice the current of the half rpm/V 16 turn motor to produce the same amount of torque. Doubling the current passing through the winding resistance will increase resistive power losses 4 times (P = I^2 *R), so while the ½ rpm/V motor will have 4 times the winding resistance, only having to supply half of the current means that both motors have the same amount of resistive power loss.
Both motors will also consume the same amount of electrical power to spin a load at a given speed: As electrical power is simply the product of voltage and current, the 8 turn motor will have to draw twice the current, but at half of the voltage, resulting in the same power consumption.
You do get a slight efficiency or weight gain from using a higher voltage system, but not thanks to motors. By doubling the voltage, you will halve the current, so you can reduce the cross sectional area of all wiring between components by three quarters while maintaining the same amount of resistive losses in said wiring.