A buck switching regulator has two phases of operation, that it switches between. For example, below are the important components of a 5 V in, 2 V out buck converter.
simulate this circuit – Schematic created using CircuitLab
In the first phase, the input is connected (usually by a MOSFET) to the inductor, which is connected to the output. The same current flows in both, while the current builds up in the inductor due to the positive 3 V across it.
In the second phase, the input is disconnected, and the input of the inductor is grounded (either by a driven MOSFET, or less efficiently by a diode). Energy storage in the inductor means that output current continues to flow, while the current falls in the inductor due to the negative 2 V across it. Obviously no input current flows during this phase.
If we assume that we switch fast enough that the inductor current 'doesn't change too much', then we can see that output current is always flowing, whereas input current only flows during the first phase. The average input current is the switching duty cycle lower than the output current.
With a buck regulator, it's not so much 'where does the extra output current come from?' It is more 'why is the input current lower?' It's because the inductor current does not flow all the time from the input, but does flow all the time to the output.
If the output current was 250 mA in this case, the input current would be 250 mA flowing for only 40% of the time, or 100 mA on average (lossless case).
Switching converters are always equipped with large enough input and output capacitors, so that the supply and load only see the average current, not the instantaneous switch or inductor current.
How would losses increase the input current? Resistance in the inductor and switches would add voltage drops to the inductor voltage, which would increase the duty cycle required to maintain the output voltage, so it leads to a higher average input current.