First, it's absolutely expected that the output current and input current (averaged over a switching cycle) are not equal in a switching converter. If the currents were equal, the efficiency couldn't be any better than a linear regulator's.
Now, let's look at a simple buck regulator:
When the switch (Q1) is closed the input current does indeed go into the load. But part of it also goes to recharging Cout, whose voltage has drooped during the "off" part of the cycle.
When the switch is open, the load still receives current, but it's supplied by D1 and Cout.
So there's no concern that by not drawing input current during part of the cycle you might fail to provide power to the load. It's just part of how a buck converter works.
Would a large capacitor on the input of my step-down converter do the trick?
A large capacitor (Cin in the schematic) won't change the fact that when the switch is open, no input current is drawn.
What it will do is, when the switch is closed, allow much of the input current to come from Cin instead of from the upstream voltage source. This current will be flowing in a relatively small loop, and so not produce so much EMI issues as if it had to flow from the upstream source, however far away that might be.
It also means that whatever inductance there is in the lines from the upstream source to your circuit won't cause Vin to droop during the "on" part of the switching cycle and interfere with the converter's operation.
Edit
I realized you're worried about the peak current drawn during the "on" part of the cycle being higher than the PoE can supply.
Yes, a larger Cin will help with that, by smoothing out the current drawn over the switching cycle. But basically any capacitance on the load side of the PoE will also help.
The choice of operating frequency and L1 value will also affect the peak current draw on the input.
I will try to answer your points assuming I have read them correctly.
1) The driver for the buck circuit will generally need some kind of regulation, but this can be a low power regulator. You then use the buck circuit to step down your supply voltage and regulate it, this will provide a lower stable voltage with more current for your load.
2) If your have either a dynamic supply voltage or load then the buck converter will need to be regulated to maintain a stable voltage. This is achieved by varying the duty cycle to the power MOSFET and feedback is required to the driver circuit to determine how the duty cycle is adjusted. You can use voltage feedback and then some closed loop control circuit to alter the duty cycle pending on the feedback signal.
To summarise a buck circuit is a good way to step down and regulate a large voltage to a smaller voltage, this can then provide power to a small, medium or large load. Additional regulation is generally required for control circuitry such as a microcontroller and driver circuit, but these circuits generally have very small power requirements.
Best Answer
You can use a fast op-amp to amplify by 2 and feed it with the output voltage (say 1.25V). The output from the op-amp feeds the reference input and if you do this correctly, the switcher will force the pulse width down until there is 1.25V on its reference pin. This means the output MUST be at 0.625 volts.
In other words you have conned the switcher by using a gain stage. There can be stability issues so use a fast (possibly 10MHz) op-amp like the AD8605 and be prepared to add some filtering capacitors.
You can also do this by adding a DC offset into ref signal. So, take the output voltage and pull it up with say a resistor and diode. This raises it by ~0.7V and will force the switcher into producing an output that is 0.7V lower on the output. No gain involved here so stability probably won't be an issue.