If an electron has 0 electric potential after passing through a resistor, how does it flow to the other terminal
Firstly, remember that the current is a migration of electrons jostling and pushing each other in the general direction of the positive terminal rather than each electron shooting along the wire from negative to positive. Pushing an electron in at one end of the wire pushes one out the other end. This happens close to the speed of light even though the individual electrons are moving much, much slower than that.

Secondly, remember that the potential difference causes current to flow. Let's use another dodgy water analogy.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Yet another water analogy.

Figure 1 shows a water tank with a down-pipe of resistance R1, a horizontal pipe which is relatively wide bore and has no resistance and another down-pipe of resistance R2. Water flows in the circuit due to the pressure difference between the top of the tank and the outlet (at the bottom of R2).

Question: There is no height difference (potential difference) across the horizontal pipe so how can the water have the energy to flow from left to right?

By now the answer should be fairly obvious. It's getting pushed by the potential difference in other parts of the circuit.

In the electrical circuit of your question the electrons can't all just pile up at the end of the resistor they have to continue to return to the battery. If they didn't the battery would get out of balance and refuse to supply any more out the other end.

I hope this helps.



Comment:


  However if a circuit only has 1 resistor, electrons which pass through the resistor is not in between 2 points of different resistance. The electrons and the battery have no potential difference between them any more after passing through the resistor. So why will the electrons flow?








simulate this circuit

Figure 2. A single resistor water analogy.

Again, pushing the dodgy water analogy a bit further, we have two horizontal big pipes and one vertical small pipe, R1. We all know that the tank will drain despite the horizontal sections. What's driving the current is the potential difference between the top of the tank (battery +) and the open end of the pipe (battery -).

Keep asking questions if it's not clear. You are close to understanding but judging by your other question, Electric potential in a cicuit, you've got static and current slightly jumbled in your thinking.