transformer
If a transformer with only one secondary winding, the reflected impedance from the secondary winding is
$$
Z_{p,eff}=(\frac{N_{p}}{N_{s}})^{2} \times Z_{s}
$$
This is a well known equation. But what if there are two secondary windings with turns \$N_{s1}\$ and \$N_{s2}\$ each? The simplified circuit is as below (Note: I put 4 meters in the circuit, and denote the readings beside them):
What's the reflected impedance to the primary?
A high impedance meter in most cases will give more accurate readings because it does not draw much power from the circuit under observation, but it can give misleading readings when the circuit produces high voltages without a load. What you are observing is the result of an overly sensitive meter combined with inductive and capasitive coupling. If you put a load on your transformer, you won't notice this.
What you may be getting confused about is the "ideal transformer" in the equivalent circuit. You should not regard it as having any magnetic qualities at all. Try and see it like this: -
Whatever voltage you have on the input to the ideal transformer, \$V_P\$ is converted to \$V_S\$ on the output of this "theoretical" and perfect device. It converts power in to power out without loss or degradation such that: -
\$V_P\cdot I_P = V_S\cdot I_S\$
The ratio of \$V_P\$ to \$V_S\$ happens to be also called the turns ratio and almost quite literally it is on an unloaded transformer because there will be no volt-drop across R3 and L3 and only the tiniest of volt-drops on R1 and L1.
This means you now have a relatively easy way of constructing scenarios of load effects and recognizing the volt drops across the leakage components that are present.
The equivalent circuit of the transformer is really quite good once you accept that the ideal power converter is "untouchable" and should just be regarded as a black box. For instance you can measure both R1 and R3 and, by shorting the secondary you can get a pretty good idea what L3 and L1 are. With open circuit secondary you can measure the current into the primary and get a pretty good idea what \$L_M\$ is too.
Best Answer
Two "perfect" identical secondaries produce identical voltages and are exactly equivalent to one secondary winding that has the same number of turns as either of the original windings. This means that the loads can be directly paralleled and reflected back to the primary as you would a single load on a single secondary.
Another way to look at is to consider the power in and power out of a perfect transformer with two secondaries. With only one secondary connected to a load dissipating 10 watts, the power into the transformer (ignoring losses because it's perfect) is 10 watts. If you connected up a 2nd identical load to the other secondary, the power into the transformer would be 20 watts.
This would also be the same should the two loads be connected in parallel to just one secondary winding or both windings (respecting phase directions) were wired in parallel and connected to the two loads.