They're training you very old school, presenting loads as if they are basically sinusoidal, big clunky old motors and ballasts. I find the "watts vs VA" question to be better served by modern electronic examples.
You're building a holiday display for the village. The mayor's wife bought 400 LED Christmas light strings at deep discount. They are wired as a single series string of ninety LEDs and a diode to guard against reverse current. They're only on for half the power cycle (they shimmer) but this is very typical of modern LED Christmas lights. Meaning they are on for half the AC cycle. The plugs are polarized and modifying them is off limits.
For math simplicity, power is British 240V AC single-phase. Each string has a 6 watt draw. (this is our WATTS unit). The 400 strings draw 2400W together or total 10 amps. Remember, their power draw looks like this because the diodes are all facing one direction.
Now, you are sizing the generator to drive this load (and no others). It's an ideal world, so a 2400 VA rated generator can carry a 2400 VA load. *
Will a 2400 "watt"/VA generator do the trick? Hint: Will a 10A fuse do the trick? **
NO! It won't! Because the LEDs are drawing 2400W on average but only drawing power half the time, that means they are drawing 4800W or 20A when they draw power.
The fuse would blow because it would run hot. * As would the generator windings, which is why we can't use the generator's kinetic energy as energy storage.
Without any means of time-shifting, the generating capacity at the bottom half of the sinewave is wasted and useless.
The load may only use half the sinewave, but we must generate the whole sinewave. Thus, our generator must generate the full 4800 VA.
And that is how I think of the VA unit. Watts is the part of the sinewave you actually use. VA is the whole sinewave you must generate.
* Hint, hint.
** By the way, a sidebar on thermal matters. Work it through Ohm's Law and you see that for a given resistance and ampacity, thermal rise is proportional to the square of current. Thus, twice the current half the time == 4x the heat half the time == 2x the heat in net for the same watts. Really. Crunch the numbers, you'll see.