Thanks for the explanation, the situation is clearer now.
This is how I would tackle the problem:
Your Dev board obviously has on board rectifiers so it can accept an AC supply but you are feeding it DC anyway. This means the rectifiers are not really serving any useful purpose and the ones on the ground side of the bridge are responsible for the boards ground being 0.7V above the power supply ground. If possible, short out the bridge rectifier so the negative side of your 12V is connected directly to the boards ground. If that isn't possible, provide a wire link between supply ground and the board ground at the boards ground terminal. This effectively shorts the diode out but externally to the dev board.
With only one ground, you don't have to worry about the voltage drop as the transistors emitter pin and dev board ground will be at the same potential and both returned to the same point at the power supply.
Switching more than one load is not a problem. You can switch as many as you like until the power supply can handle no more. Bear in mind that as well as the current being drawn by your load, you have to add the current drawn by all the relay coils as well when you work out the total load. Use a transistor to drive each relay like you do at the moment but drive the base from different signals from the dev board if you want to control them individually.
Brian.