What confused me was when he asked whether electrons (current) can travel both directions at the same time through Light 1.
Well, the answer is yes, and no. Electrons can, and do, travel through Light 1, and in fact all metals, in both directions, all the time. Unless you can manage to cool a piece of metal to absolute zero, then the electrons are wandering around in random directions all the time, much like individual water molecules are wandering about in an otherwise stagnant glass of water.
But, when we talk about electrical current, we are talking about the net flow of electrons. If we say there is a current in some direction, what me mean is than on average, electrical charge (electrons being but one type of such) is flowing on average in that direction.
There may indeed be forces acting on some component that individually try to move current in opposing directions, but what's important is the net force and the resulting net current, much like two teams pulling on a rope in tug-of-war, or two sumo wrestlers pushing against each other. The net force is what determines the motion.
If V1 were to become an open, then V2 would have to supply power to both Light 1 and Light 2 in series. (Pretending for the moment that these lights would simply be more dim.)
Given that current from V2 travels through both lights in this series circuit (with V1 open), how is it that the presence of V1 causes current flow from V2 through Light 1 to cease?
It doesn't. The current in each light doesn't "belong" to either V1 or V2. Who knows, or cares, where each charge carrier came from? Consider what I just described about the electrons wandering around from thermal noise. Also, consider that their movement due to electrical current is relatively slow, and you will see that this is an irrelevant question to ask.
Here's another way to think of it. An open circuit, by definition, allows no current. It's an infinite impedance. A voltage source, on the other hand, passes whatever current is necessary to maintain its voltage. If something else wants to push more current through it, and that won't change the output voltage, it won't resist at all. Thus, it's a zero impedance. V1 does as much to impede the current in V2 as a short circuit would do. But it also must exert some force on the charge in the circuit to supply additional current so that it can create an additional 1.5V of difference across Light 1 and Light 2.
That is, V2 has to push all of the current for Light 2, but it only has to push it over half the electric potential difference (voltage), because V1 is pushing it the other half of the way, in addition to pushing the current needed to power Light 1.
Further reading: Thévenin's theorem, especially the part about "Replace voltage sources with short circuits", and Kirchoff's circuit laws.
Some cheap chargers don't use a regulator on the output side, monitoring the output voltage, and feeding back this information to the primary via an optocoupler. Instead, they regulate based on what the "see" on the auxiliary winding on the primary side, saving the cost for the optocoupler. On the aux winding of a flyback converter, you get a very approximate information about the voltage on the secondary winding, and sometimes, ringing and spikes dominate the picture instead of the theoretically ideal reflection of the output voltage. It could be that under mid- to high-load conditions, the ringing on the aux winding becomes less and the regulator increases the power just because it receives a decreasing ratio of the output voltage.
If this is the case, the error actually would occur at light loads (because this is where the ringing might be severe), but is compensated for by decreasing the overall regulation such that the voltage is somewhat within the specification over the entire load range. The details would depend on what the snubbers are optimized for, for example.
Cheapo USB chargers are sometimes reduced to an absolute minimum of components because the cost pressure is dramatic. Muntzing is anything but an old-fashioned trick of the trade.
Searching for "primary side regulation in flyback converters" might lead to some background info like this article.
Best Answer
The current control is implemented by the power management logic implemented in the phone. In particular, according to the Mi 4i FAQ, your phone uses Qualcomm's original QuickCharge 1.0, which consists primarily of two components: AICL = Adaptive / Automatic Current Control, and APSD = Automatic Power Source Detection.
APSD incorporates BC1.2 = Battery Charging Revision 1.2, an extension of the USB standard that allows chargers to communicate that they can deliver higher current, e.g. by shorting the D+ and D- lines.
AICL includes heuristics for automatically detecting the current limit of the power supply by slowly stepping up the input current (e.g. in 25mA steps). Usually this is limited to about 2A.
You ask in the comments if you can safely charge your phone using a DIY charger using a 5V regulator. Generally this should work fine as long as you include the appropriate BC1.2 signalling to the phone. To do that you can either construct your own fast-charge cable with the D+ and D- wired as needed (see below), or you can buy various fast-charge USB adapters, e.g. search on "CW3002" on eBay to locate adapters using a common CellWise CW3002 controller.
Below is a bit more info on the technical details, first on on AICL, from the NXP (freescale) BC3770 datasheet, and second on BC1.2 from Maxim's Overview of USB Battery Charging Revision 1.2 and the Important Role of Adapter Emulators. Now that you know the appropriate buzzwords you should be able to dig deeper if need be.