Potential Difference
A battery is connected to a light bulb. Charge flows round the circuit. It experiences resistance in the wire, light bulb and battery. The charge has to do work to overcome this resistance. The energy for this work comes from the battery.
Consider points B and C, on two sides of the light bulb. Charges flows from B to C through the light bulb. (It is actually electrons that flow from C to B, but we think of this as being the same as positive charges flowing the other way.)
We use the idea of potential difference to tell us how much work is needed. Suppose that 2 J of work is needed to move 1 C of charge from B to C. Then we say that the potential difference between B and C is 2 volts, or 2 V. That is, 2 V = 2 joules per coulomb, of 2 J/C.
The wires are often made of copper, which has very low resistance. Very little work has to be done by the charge when it goes through copper wire. Suppose, just for example, that a total of 0.01 J of work is done when 1 C of charge goes through the wire from A to B, and from C to D. So the potential difference for the wire is 0.01 J/C = 0.01 V.
There is also resistance inside the battery. This is called internal resistance. Suppose that when 1 C of charge goes from D to A through the battery, 0.5 J of work is done.
If we add up all the work done to move 1 C of charge through wire, light bulb, and battery, we get 2 J + 0.01 J + 0.5 J = 2.51 J.
The energy for this work comes from the battery. It is the work that the battery has to do to move 1 C of charge around the circuit once. This work is called the emf. So in this example, the emf of the battery is 2.51 J/C, or 2.51 V.
This means that we can add up the potential differences across different parts of the battery to get the emf of the battery: 2 V + 0.01 V + 0.5 V = 2.51 V.
To summarise:
Potential difference between two points is the work done to move one unit charge from one point to their other.