NOTES: If you want complete optical isolation, connect "Vcc" to Arduino +5 volts but do NOT connect Arduino Ground. Remove the Vcc to JD-Vcc jumper. Connect a separate +5 supply to "JD-Vcc" and board Gnd. This will supply power to the transistor drivers and relay coils.
Having the seperate 5V go to the JD-Vcc means the Opto-receiver and the Relay are totally isolated from the Arduino Vcc. Overkill for sure, but the design provides for that.
If you want complete isolation, you will have seperate 5V supplies, one for the Arduino and one for the relay power.Then you would connect an arduino output pin to the cathode of the drive LED, with VCC to a current limit resistor to that anode.On the other side, you have pin 1 of the jumper field goes to the pin that goes to Pin 4 of the optoisolator.If you only have 1 wallwart, leave the jumper in. Having an optoisolatar and the relay seems like overkill to me. Most applications just drive the transistor to drive the relay.
What are you switching with the relay? If mains voltage you need opto isolation, can you post a link to that relay module?
So why a 3.3 V Arduino Pro Mini? Pro Mini itself is perfect, but 3.3 V an odd choice. You are clearly not saving power!
Note of course, that the 3.3 V Pro Mini outputs switch between 0 and 3.3 V, while the relay module VCC comes from 5 V. This is absolutely correct - the relay module has two LEDs in series on its input with a combined voltage drop of 2.6 V or so, so it may not reliably switch from 3.3 V alone and will certainly not conduct at the difference between the 3.3 V HIGH and 5 V, but when the outputs go LOW it will see the full 5 V and operate reliably.
while the relay module VCC comes from 5 V.