The more important thing is the environment is completely different: a mains transformer is connected to a low impedance, constant frequency, voltage source. The transformer is designed to saturate not much higher than nominal voltage (or lower than nominal frequency). There is little or no inrush at the instant power is applied to the transformer; the inrush surge occurs after some delay (i.e., at least 8ms, but more likely closer to 20ms).
In a typical (current mode) SMPS, the supply voltage or frequency is effectively variable, or the source impedance is effectively higher. To see why this is, we must cover several relations:
* Voltage in RMS, is the root of the mean of the square of the instantaneous voltage. That is, for a signal v(t), take v(t)^2, then average it over some period of time; then take the square root of the result. In calculus, we can write:
\[ V_{RMS} = \sqrt{ \frac{ \int_0^T v(t)^2 dt}{T} } \]
The integral sign means to add up many tiny slices (of width \$dt\$) of the expression within, over the range specified by the indices (i.e., 0 to T; for a periodic signal like a sine or square wave, we can set T equal to the period).
* The RMS of a "pulsed DC" wave (i.e., a load driven by a switch to a DC source) is therefore \$V \sqrt{D}\$, where D is the duty cycle of the wave.
* In a current-mode controller, for a given current setpoint, and for an inductive load, the on-time is set by the inductance:
\[ V = L \frac{\Delta I}{\Delta t} \]
In DCM, current starts at zero so \$\Delta I\$ is also \$I_{pk}\$. So the on time \$\Delta t\$ is proportional to the set current and the inductance.
Which means, as the load impedance (taking inductance as approximately impedance) decreases, the switch duty decreases, so the load voltage decreases. In other words, the source impedance is significant.
We can't simply measure the source impedance, because the circuit is very nonlinear (there's a comparator setting switch time), and it's time-domain not AC-steady-state. We have to be very deliberate in defining what we mean by "impedance" in this case. Following the above definitions, we could indeed end up with an unambiguous proportion having units of ohms; it wouldn't be terribly useful though, so it will suffice to say that we can, and that it more-or-less follows the usual rules of impedance, at least within the limits of our definitions.
In short, it doesn't get inrush because it's not a low impedance mains source.
And we can do a similar exercise for the variable-frequency case, where applicable (e.g., the MC34063 and ilk).
Regarding power dissipation, it is usually divided into two aspects: conduction loss and switching loss. Roughly speaking, we can calculate conduction loss as Irms^2 * Rds(on) (Irms is calculated same as above, incidentally!), and switching loss as 0.5 * Vpk * Ipk * t_f * Fsw (draw the V and I waveforms during switching, to see a geometric proof of why this might be so; hint: what figure has area "one-half base times height"?). A lot of subtle dynamics is going on with modern transistors (mainly because of their highly nonlinear Coss vs. Vds relationship) which tends to dissipate less switching loss than this would suggest, although it can be increased in some situations instead (hard switching).
Keep drawing circuits and waveforms, and running simulations; this won't (can't?) come instantly, but through careful study you can understand -- and prove -- everything that's going on here.
Tim