So this is my recommendation (unfortunately I cannot make any drawings at the moment):
For the output (the side where the fans are connected)
- Connect the red wire of the fan to its power supply. That means to the +5V supply (red wire) from the microUSB board.
- Connect the black wire of the fan to the COM pin of the relay (that would be pin no.2)
- The NO pin of the relay should go to the GND from the microUSB board
- And the NC pin of the relay should be left floating (or connected to the +5V from the microUSB board)
Having the above connections you achieve the following:
- When the relay is not activated (that means the COM is connected to NC) the fan sees on its wires +5V and floating (or twice +5V), so no current is flowing and the fan is not working.
- When the relay is activated (that means the COM is connected to NO) the fan sees on its wires +5V and GND, current is flowing and thus the fan is working.
Mind that usually the max. current from USB is around 500mA, so make sure that the total current needed by your fans (if you supply all of them in parallel from the same usb port) does not exceed this value.
Then for the input side of the relay board.
- The inX pin of the relay should go to a GPIO of the Pi
- The GND pin of the connector (left bottom side of the board) should go to the GND of the Pi board
- The VCC pin of the connector (left bottom side of the board) should go to the +3.3V power supply of the Pi board.
- I would connect the +5V to the JD-VCC pin. Also don't forget to remove the jumper on the right side of the relay board.
As you see in the diagram in the link you posted from another question, the JD-VCC is applied directly to the input of the relay and there is an optocoupler between the inX pins and the relay input. Since in your case you probably don't really need the extra isolation, connecting the JD-VCC to the +3.3V (either with another wire connection or with the jumper on the board) would probably also work. But I'm not sure what input supply voltage the relays used there are designed to work for. Chances are (but again perhaps I am wrong) that it will not work with the +3.3V supply.