Voltage in a circuit? 
The voltage of the battery signifies the difference in voltage between the positive and negative terminal

What does this mean?
The definition of voltage difference I'm familiar with is the amount of potential energy charge one coloumb of charge would undergo from point A to point B.
At first I thought the quote was saying the voltage of the battery signifies the difference in potential energy one 1 C of charge would have if it were moved from the positive to the negative terminal. 
But my interpretation of batteries is that they "push" electrons into the wire which push the electrons in front of them ahead which push the electrons in front of them ahead and so on.
How can I connect the two concepts of voltage being the difference in potential energy a charge would have if it moved from the positive to the negative terminal with the "pushing" power of a battery?
 A: You can reconcile both trains of thought by reconsidering your thoughts about pushing:-    


*

*For DC circuits without changing magnetic fields, the voltage is the energy gained or lost per unit charge in moving from one position to another, say from the positive to the negative terminal of the battery.   

*What a battery does, is that it creates certain junctions which intrinsically have a potential difference between them even in equilibrium. The battery, through chemical reactions, maintains this potential difference. The potential difference between the two ends of the battery is the culmination of all these junction potentials within the battery which is maintained by ongoing chemical reactions. So the battery's job is to only maintain a potential difference across its terminals, and the rest of the drama plays out on its own.     

*Now, if we connect the two terminals of a battery by a conducting wire, the wire now has its ends at different potentials, that is, there is a potential difference across the ends of the wire. This potential difference sets up an electric field within the wire from the positive to the negative($\vec E=-\nabla V$) terminal. This field is what does all the pushing. This field propels the electrons towards the positive terminal.  

*Since the electrons keep colliding with the surrounding metallic kernels, they never acquire a constant acceleration, but they acquire a constant drift velocity with which all the electrons slowly edge towards the positive electrode. When the electrons collide, they lose energy as heat. Therefore, when electrons having unit electric charge come from the negative end to the positive end,colliding along the way, the amount of energy dissipated as heat (or any other way) will be equal to the potential difference between the two ends of the wire, which is the terminal voltage of the battery. All what the battery does is that it carries on chemical reactions to maintain a constant potential difference between its ends.   

*The condition when the circuit is setting up is distinct from the condition in steady state. When the circuit is starting the electric field is still not established along the wire, i.e all the electrons along the wire do not feel the push of the electric field. Here, the end closest to the terminal "feels" the potential difference first and starts moving. As only a small part of the conductor has a current, there will be unbalanced charges which make the next part feel the potential difference, and so on until the other end of the wire. This is similar to the kind of pushing you describe. But all of this happens to fast to matter in most circuits and the condition after the circuit is setup is described above.

