Do you have any part of pull chain left in the old switch, even if you need to use tweezers or something to use it? If you do, then you could use a continuity tester or volt-ohm meter and test how the switch is working when the chain is pull through its cycles. That is, see which wires let voltage through (and it may be more than two at a time).
To me, "L" should be the hot voltage in to the switch, while 1-3 should be power to the fan.
If you can not test the switch directly, perhaps this article will shed some light.
http://www.ceiling-fans-n-more.com/ceiling-fan-pullchain-replacement-and-repair.php
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...
Best Answer
In general, yes slower speeds use less power. The fan motors typically are synchronous AC motors and use different windings or capacitors on the motor to create different speeds. There isn't a "dimmer" circuit or resistor speed control when you use the pull chain switch.
If you try to use a dimmer switch instead to change the speed that isn't designed for fans the quality of the power (power factor) can drop which can lead to additional inefficiencies.