Electronic – DC-DC Step-Down (buck) converter 5V 10A 50watt, not understand how to size the supply

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.

You said you measured the 1.3 Amps across Vin and GND. This is NOT the way to measure the current capability of a power supply - it does give you the short circuit current, but that is usually not a useful value. If the existing power supply claims to be 12 volt and 900 mA, you should believe that current rating, and not attempt to draw more currrent.

To measure current, you must connect your meter in series with the circuit - you break the circuit to do this.

You should measure the actual current drawn by the 12 volt load. If it is less than 900 mA, then the difference is the current you have available to power your step-down converter.

As others have said, the 300 mA rating of the stepdown converter is the maximum it can supply, not what it will actually draw. Since the stepdown ratio of the converter is 5/12, we can expect the current drawn from the 12 volt supply to be a bit more than 5/12 of the load current - perhaps 35 mA. (As a rough approximation, we can assume that the power into the converter equals the power out, plus some losses in the converter.)

The total current drawn from the 12 volt supply will then be whatever you measure for the scale, plus the 35 mA or so for the step-down converter and 5 volt load.