Don't assume I'm any good at this, think I said that I'm pretty bad with analog stuff. I'm going to go on a bit here, but I'm trying to explain my thinking in case you or anyone can spot If I'm going wrong somewhere.
So, I think we both agree about the function of the zener diode: when reverse biased then a zener won't conduct any current until the threshold is reached, at which point it will start to conduct however much current is required to keep the voltage from rising any higher, until it overheats. So in your example, the 3.6V zener can have anything from 0V to 3.6V across it, but nothing higher.
In plumbing terms, it'd be a sort of pressure activated valve (a bit like a relief valve). Think of something like those ketchup sauce bottles with the rubber seal, where you have to squeeze the bottle quite hard before anything starts to come out, and then will suddenly stop and seal up when the pressure drops (doesn't quite work as an analogy because the pressure to start the flow is higher than the pressure to maintain the flow, but I don't think it's too far off).
That's what I mean by the zener dropping 3.6V. If the voltage/pressure across it is less then the threshold, then nothing goes through. Once the voltage/pressure is above the threshold then potentially lots can come through. But if there's some blockage/obstacle/resistance further along, then the pressure will start to build there. That decreases the pressure difference/voltage across the zener. If the pressure across the zener drops too far then it will stop conducting.
Now, I think where things get a bit different is exactly how the zener fits in to the circuit. In your example the battery provides 9V: that's the pressure. The zener threshold is 5.1V. So without the resistor then lots of current would flow through the zener and it would break. So there's the 330ohm resistor. Current flowing through the resistor will lose voltage. What will happen is that enough current will flow through the resistor until the resistor output is at 5.1V. If any more current flowed then the output would drop and the zener would stop conducting. If any less current flows then the voltage across the zener would try to rise, which would mean more current would flow through the zener. So that means there's a fixed 5.1V across the zener, which can be used as a 5.1V supply to other things (up to about 15mA).
In that example the zener is in parallel with the load (the load goes either side of the zener). But on your motherboard, the zener is in series. I've tried to sketch it here:
On the right you can see the parallel version, similar to your example, but showing the battery to be charged. Voltage and current (Power) come from the top and through a 10ohm resistor, and there's then a zener in parallel with the battery. When the battery is empty then it will have 0V across it, so current will flow through it, but not the zener. As the battery charges up then the voltage will increase. Eventually the voltage will reach the zener threshold, at which point current will flow through the zener instead, and the battery will stop charging.
But that's not how the motherboard is wired up. It's connected like the left hand side, in series. Assume the motherboard supplies 5V and the zener threshold is 3.6V. Power comes from the motherboard, through the resistor, through the zener and then through the battery to ground. If we assume the battery is empty then it's at 0V on both sides. If we assume that there's no current flowing then there's no voltage drop across the resistor. That means all of the voltage will be across the zener. The motherboard supplies 5V, which is above the 3.6V threshold, so current will flow. As more current flows, the voltage across the resistor will increase, so there will be less voltage across the zener. Too much current will mean too much voltage is lost across the resistor, which might reduce the voltage across the zener below the threshold, and conduction would stop. So what will happen is there will be a steady state of just enough current flowing to keep the voltage across the zener at 3.6V.
But, over time the battery will charge up, and the voltage on the +ve side of the battery will increase, which increases the low side of the zener. To keep conduction happening then the high side must also increase. Which means the voltage across the resistor must decrease. Which means less current flows and the charge rate slows down. Eventually (after infinite time...) the battery will be charged up so that the voltage across it is 1.4V. The zener will have 3.6V across it. Which leaves 0V across the resistor, so no current will flow through the resistor. If the battery could charge any more then the voltage across the zener would drop beneath 3.6V and it would stop conducting. So no current flows and the battery stops charging.
For the battery to charge any higher then the zener needs a lower threshold. If the zener had a threshold of, say, 1.4V, then the battery could charge to 3.6V. But the lowest common zener threshold voltage I can see is 1.8V. Which leaves just 3.2V across the battery, which isn't enough.
So I think we're missing something. That diode D3 is almost certainly not a zener diode. And yet it appears to be the only way for current to flow in or out of the battery. But to charge the battery that means current has to flow backward through the diode, which isn't possible unless it's a zener diode, in which case it needs a threshold of at most 1.4V, which don't seem to exist.