Comparing those tables: Note that the speed switch in the circuit you show isn't using L.
A: L+2+3
B: L+1+3
C: L+1 (Maybe this is L+1+2 ???)
D: L+1+2+3
0: No connection (or no connection to anything but L)
1: 2+1 (possibly plus a connection to L)
2: 1+2+3 (possibly plus a connection to L)
3: 2+3 (possibly plus a connection to L)
Making them correspond with each other...
C is obviously equivalent to 0.
D is obviously equivalent to 2.
That leaves us with
A: L+2+3
B: L+1+3
1: 2+1 (possibly plus a connection to L)
3: 2+3 (possibly plus a connection to L)
We can make those match if we relabel the connections. If we just swap the labels on your terminals 3 and 2, then
A is equivalent to 3
B is equivalent to 1
If we renumbered them all (your 2 is their 3, your 3 is their 1, your 1 is their 2), then
A is equivalent to 1
B is equivalent to 3
Pick whichever you prefer; one will switch off-high-medium-low-off, and the other will switch off-low-medium-high-off.
As far as theory goes: I'm not sure either, but let's see what I can do with it.
3 (2->3) appears to be "slow" because power flows through the right half of the bottom coil, and then through the side coil, in series. More resistance, less current flow, less power.
1 (2->1) appears to be "fast" because the left side of the bottom coil, and the side coil, are powered in parallel. Both get the full house-current voltage applied across them rather than the reduced amount of power they got in series.
2 (2->1 and 3) is the tricky one. I am far from certain, so DON'T take my word for it. But I think what's happening here is that, since the middle and right sides of the bottom coil (1 and 3) are now connected to each other, that loop has a current induced in it by the motor's moving magnets, which creates a countering magnetic field, which acts as a magnetic brake to slow the motor... so fast with a bit of braking equals medium. Seems like an odd solution, but if I'm remembering my freshman Physics at all correctly it might actually be a reasonably efficient solution.
You might want to run this by the physics discussion, to get someone with more recent memory of electrodynamics to check and/or correct that last paragraph.
Gopher baroque...
Is this approach correct?
Basically, yes.
1mm2 twin and earth cable (I presume that is what you are proposing) could carry up to 16A, depending on where it runs, so 1.5A is well within the headroom (assuming that you aren't talking such long runs that voltage drop becomes an issue).
For lighting, however, I tend to use 1.5mm2 in preference to cable 1mm2.
Best Answer
A quick search indicates that acquiring 3 phase power from a local utility may be prohibitively expensive. The article I read suggests that such a resource will carry an installation expense in areas not rated for commercial service and may also carry a minimum charge on the utility bill.
If your objective is occasional use of heavy equipment such as a machine tool, it may be more practical to use a device called a rotary converter. You would want to ensure your selected device has the necessary capacity for your expected load maximum, with a bit of reserve tossed in for good measure.
If only a single machine is going to be using 3-phase power, another device known as a VFD would be of value. A VFD will provide variable speed to an otherwise single speed motor as well as other features one can determine by searching for that term.
There are plans online to build your own rotary converter. A friend without internet access, many years ago, connected two motors together via belts, powering one with line current and used the gizmo to run a monster Czech lathe in his shed. He had to "kick-start" it periodically by kicking one of the pulleys, but it worked. I'm sure a commercially produced version would not require such manual intervention.
The above assessment was performed based on a US installation. Other countries may have different circumstances.