What does voltage output from an SMPS look like when inductor is saturated

You're not really missing anything.

A switching power supply converts DC to DC by first converting AC to DC, then to AC, then back to DC:

• The input rectifier and filter converts the low-frequency AC (50-60 Hz) input into DC
• The chopper converts the DC into high-frequency AC (usually several hundred kHz)
• The output transformer steps the high-frequency AC up or down and passes it to the secondary
• The output rectifier and filter converts the high-frequency secondary-side AC into DC

So yes, at it's heart, there is AC transformation going on. The frequency of the AC is made much higher to allow for a much smaller and less lossy transformer to be used in the conversion process.

A push-pull converter, as you've described in your question, is AC to AC at it's heart.

The action of the two switches chops the DC input bus into AC by applying the input to the transformer with alternating polarity. This AC waveform is converted to the secondary through the transformer, then rectified back into DC. Per the article,

The transistors are alternately switched on and off, periodically reversing the current in the transformer.

Periodic reversing of current = alternating current = AC, just to be explicit.

Note that the image shows only part of the power train. Here's a more complete representation of a push-pull converter:

This shows an output LC filter, which is needed to make proper DC on the output side.

Of course, transformers only work with AC. You cannot apply DC to a transformer and expect any output, since transformers only "work" by a change in primary current generating a change in flux which is coupled to the secondary and generates a change in voltage. Change = AC.

When it comes to DC, all you can really do without some intermediary AC stage (be it a transformer, an inductor, or even flying capacitors) is dissipate power to lower the voltage. This is what linear regulation is - dissipative.

Also, you can never increase DC voltage without some form of intermediate AC stage.

Note the word "dissipative" - that's why in the vast majority of cases (when the power levels are not trivial) a switching power supply will be more efficient than a linear solution.

You have several things to worry about in this setup. A schematic of what you have exactly would have helped to target the advice more to the exact problem, but here goes:

First of all, you may not necessarily need insanely large capacitance, but you need capacitance with a low ESR (Equivalent Series Resistance). If your 470uF capacitor has a large resistance the regulation of the buck converter will improve and the peak current capacity of the capacitor will increase, solving the problem from two ends. That's not to say you may not need a higher value if you don't want a dip at all.

Secondly, you need a low ESR capacitor on the input of each converter stage as well, to make up for resistance in batteries and/or wiring, so the converter has a local store of energy to pull from when the output suddenly needs 1 or 2 ampere for the motor's cold start.

From the datasheet I expect the response of the regulator to be quite sharp, so it should be searched in the capacitors first. If you cannot get or afford a very low ESR capacitor, you can combine two or three lower value capacitors, as their ESR will be used in parallel and as such the total ESR of the capacitor-group will be much lower.

The second picture does seem to show a retry/restart of the motor and the regulator's response to that transient current. It's likely the servo has a little margin when the power to the motor falls off. More advanced servos are better tuned to prevent from showing behaviour like this.