Your RLC meter assumes that the parameter you select is the only significant attribute of the item being measured, so when you tell it to measure R it assumes that the stepper motor coil is a pure resistor with no inductance or capacitance. In reality the coil has significant inductance which will increase the reading. At 10kHz the coil's impedance is mostly inductive, so the L reading should be fairly accurate but the R reading won't be.
For PWM current control the inductance value isn't critical - it just has to be high enough to keep current ripple down to an acceptable amount. However the resistance value is critical, as it determines the average current through the coil. At high frequencies the resistance will increase due to skin effect, but in the frequency range you intend to use and with the small diameter of the coil wire, its AC resistance should not be much higher than your DC measurement.
Ignore the RLC meter's AC resistance measurement. The DC value should be close enough. Use the lowest available PWM frequency that still maintains continuous current flow in the coil at the lowest PWM ratio you intend to use.
Firstly, there is an error in the dot notation - scenario 4 provides the highest inductance yet the dot notation implies that if the two inductors were perfectly coupled, the net inductance would be 0. Because scenario 4 gives the highest value of inductance it can be concluded that it really has the dot notation of scenario 3.
Secondly, you have not considered that the two inductors may not be 100% coupled.
Next is to work out the coupling and a bit of math in my head tells me it's about 70%. Individually each winding has about 600 uH and in series aiding this rises to about 1800uH. If the two windings were 100% coupled they would produce a total inductance of 2400uH when connected in series.
So if 70% of each winding is perfectly coupled then the total inductance is: -
(4 x 0.7 x 600 uH) + (2 x 0.3 x 600 uH) = 2040 uH. OK my head-guess was a little optimistic
on coupling. 50% coupling realizes an aiding inductance of 1800 uH.
When put series opposing, 50% of the coupled inductance totally cancels leaving a net inductance of about 2 x 300 uH.
Near enough.
EDIT to explain my math
The standard formula for coupled inductors is: -
\$L_{EQ} = L_1 + L_2 + 2k\sqrt{L_1L_2}\$ and, when both inductors are the same value this results in: -
\$L_{EQ} = L + L + 2kL\$ and, when k=1 (100% coupling), equals 4L
If a fraction (70%) of L1 is 100% coupled to L2, the fraction produces an inductance of 0.7 X 4 L.
The remaining uncoupled parts of L1 and L2 do not interact and are just additive i.e. (1-0.7) X 2 L.
Hope this makes sense.
Best Answer
Raising the frequency to 100 kHz could certainly mean that you are approaching the natural self resonant frequency and you will get stupidly high inductance values and sometimes, if the frequency is too high it will behave as a capacitor and your meter displays nF or pF.
The natural self resonance is brought about by the interwinding capacitance i.e. It becomes a parallel resonant impedance.
I'm not saying definite but I am saying there's a decent probability given the size of the core. I'd trust the inductance reading at 1 kHz but by no means is 1 kHz a panacea.