Both topologies can work with either a half-bridge or a full bridge, despite in this case the 2-stage topology uses a half-bridge whereas the 1-stage I describe would use a full bridge, but this is an irrelevant difference that only affects the max power the inverter can support.
Anyways, irrespective of bridge type, what it does is effectively alternate the polarity of voltage from the DC power source on the poles of the primary coil of the transformer, at high frequency. So we would draw a PWM signal plot as a voltage on the y-axis, and time for the x-axis. The positive and negative voltage swings would be equal in magnitude, thus equidistant from the horizontal time axis.
However, if it were a perfect square wave, ie 50% duty cycle, then the integral of voltage over time for both halves of the square wave cycle, would be 0 (area over and under the graph would be equal so cancel out). The integral represents the total difference in reactive current that would pass through the primary coil. This must always be zero in any case, but during each swing, the area will temporarily be non-zero, and the magnitude of this depends on frequency. So the higher the frequency, the smaller the total current difference (per swing), hence the smaller the transformer can be. A pure sine wave would have much larger areas for each half-cycle, meaning larger current swing, which requires a much larger transformer.
So far so good.
Once one modulates a sine wave over a pwm signal, no matter what the frequency, the average areas over and below the time axis, won't be equal anymore, and they would continue adding up in the same net direction, until the complete half-cycle of the modulated sine wave, because to modulate, the duty has to continuously change to reflect the instantaneous magnitude of the sine wave, so most of the time the wave will be asymmetrical in terms of +/- voltage swings, so after each period of the high-frequency carrier signal, there will remain a net current in the primary coil that continues to sum up over successive carrier periods. So one has to continue integrating over the period of the longest wavelength to get the total swing in each direction. Hence the total area change and thus the total current swing for a 60Hz sine wave modulated over a higher frequency PWM signal, will be exactly equal to the area if one had applied the 60Hz sine-wave directly, which would obviously saturate any transformer that can't already handle a pure sine wave. Yes, as the frequency increases, the area under each individual voltage swing is smaller, but the final total current swing will still accumulate until the longest half-cycle has been considered, which is in this case the 60Hz sine wave.
What adds to the confusion, is that although both topologies result in functioning inverters, and both topologies are represented and explained and actually built in various tutorials and blogs and videos on the web, a 1-stage topology still does require much larger and heavier transformers to work, yet many of these sources actually show small transformers, which is misleading and actually incorrect. Most likely they copy their designs from each other without knowing the full story so a mistake can propagate and amplify this way just like any rumor.
Moreover, with 2 stages, each stage can operate at a different pwm/carrier frequency. Hence the 1st stage can operate at the frequency that minimizes the transformer losses, while the 2nd stage can operate at the frequency that minimizes the size of the needed output filter.
Moreover, 2 stages is easier to regulate, since as the load increases, the voltage will drop, which the signal-generator will have to counter (and vice versa as the load decreases to avoid over-voltage). In 1-stage, it means the high AC voltage has to be sensed and factored in, which is difficult to do in a stable/reliable and/or isolated manner.
However, with 2-stages, then the 1st stage produces a fixed frequency square wave (50% duty), and the transformer output passes through a bridge-rectifier and fills an electrolytic capacitor that hence stores high-voltage DC (the transformer is chosen so that this voltage always exceeds the peak AC voltage, to account later for voltage drop). There is no regulation necessary here and infact nothing to regulate, because you can't change a square wave in any way that would change the output voltage, and the high DC voltage on the capacitor after the rectifier-bridge, would remain constant independent of load, because current will pulse through as long as necessary (because it is effectively a true flat DC voltage after the rectifier) so maintains the voltage. So the regulation is instead done purely within/at the 2nd stage, hence maintaining isolation.
The 2nd-stage pwm modulates a 60Hz sine wave on a higher-frequency carrier, where the duty would oscillate around the 50% mark at 60Hz. What max/min duties it goes to, is a matter of regulation. It senses the output voltage, and when this drops, it increases the duty swing (all the way up to full duty swing), and when the voltage rises above the target voltage, then the duty swing is decreased as much as necessary. This is possible because with no load, the duty cannot and should not swing all the way anyways, because the source, the high-voltage DC, is already higher than the peak (and as mentioned must be to enable compensating increasing loads as well as attenuation in the filter).
Also, both stages can remain fully independent/isolated from each other, ie not only use different pwm frequencies, but also independent/isolated signal generators (such as microcontrollers) and logic-level power supplies.
Alternatively, the 2nd stage can keep pole "A" connected to the storage capacitor's negative pole for one 60Hz half-cycle while pwm-ing pole "B" of the output (that goes to the filter) on the positive pole of the storage capacitor, and then on the other half-cycle, it keeps pole "B" at the negative capacitor terminal while pwm-ing pole "A" on the capacitor positive pole. In this case, the duty will oscillate at 60Hz between 0 and some max value that is less than unity, and only increases towards unity as the load increases. Once it maxes out, the inverter has reached max rated power and can only give more power if the output AC voltage is allowed to drop.
How does one achieve an initial over-voltage? Well it depends on the voltage of the power source and the transformer ratio. One might choose the ratio so high that even when the batteries are near empty hence have a lower voltage, that there is still enough over-voltage at the storage capacitor that one can still provide full output rated power without any voltage-drop, but this means the over-voltage will be very high, which is risky, because if the control-loop isn't rock-solid and lightning fast, then as the load decreases (or is suddenly disconnected), then dangerous over-voltages can temporarily appear on the AC output.
Also, the Filter inductor can remain small despite 60Hz passing through it, exactly because it is blind to it and unable to react to it. The load takes care of this. Even without a load, the open terminals "react" to that. AC current still passes through the inductor, but only due to the filter capacitor, and the capacitor innately blocks low frequencies and DC.
A transformer without anything connected to its secondary, would be like an inductor/filter, and an inductor/filter with another coil wound around it, would be a transformer. So they are very similar. So without a load, the transformer is like a single inductor connected in series with the power source, without any capacitor, meaning the inductor (transformer), is itself the load, and itself has to react to all frequencies present in the source. So if the source has a 60Hz component, then the inductor would have to be much larger to be able to react to this, otherwise it saturates and a short-circuit current arises. This doesn't happen with the filter inductor because it is in series with another load. This means the constraints on the transformer are tighter than on the filter inductor, which is why one may have 60Hz pass through it but not the other, and this in turn necessitates 2 stages, ie two sets of switches and pwm signals (unless you are willing to use a transformer (inductor) with a much larger core so it can store more energy, which is equivalent to being able to react to lower and lower frequencies).
Btw... if you were to connect the hypothetical secondary coil around the filter inductor to some tertiary load, then because of the coupling, it would siphon energy out of the inductor which would be compensated by whatever is powering the filter, and it is now defacto a transformer (the output voltage would depend on the turns ratio). In this case the current passing through the filter inductor can exceed its saturation current without problem, and this would happen in any transformer anyways.