As long as both secondaries are rated for the same current, that should work fine. If they have different current ratings, you will want to take no more current than the lower of the two.
The main thing is to keep track of the polarities, i.e., the placement of the dots. With series secondaries, if you get one winding flipped, the secondaries will be out-of-phase and you won't get any current, assuming the secondary voltages are the same. If they aren't, the higher voltage secondary would drive current against the lower voltage one, but just keep the dots straight and this is not a concern.
The real hazard is when you try to put secondaries in parallel. Get the dots wrong on this setup, and the two secondaries become a short circuit. Much smoke will follow.
With the secondary open circuit, only the primary winding has effects on results and therefore the current that flows is the magnetizing current. The magnetizing current produces losses and these are called iron losses and they are partly due to eddy currents circulating in the the iron laminates that make the core. Iron isn't the best conductor of electricity and so it's resistance dissipates power due to these eddy currents.
Hysteresis loss is also measured by this test and, strictly speaking the total "loss" with the secondary open circuit is eddy current loss plus hysteresis loss (Remanence effects in the iron material leading to energy loss every cycle of ac).
Shorting the secondary tests copper losses and this is usually done at a much lower applied voltage to the primary. Because of this, iron losses are negligible and can be ignored. In this test you are measuring the I^2*R losses of primary and secondary together. If you know the turns ratio you can make reasonable assumptions about the distribution of loss in primary and secondary.
Below is the equivalent circuit of a transformer with the secondary (copper losses) referred back into the primary. This referral is done by multiplying them with turns ratio squared: -
Note component Xm - this is the reactance of primary inductance of the transformer. Xp and Xs are leakage reactances (turns of copper that don't couple magnetically).
Measurements
Watt-meter measurements will indicate the losses on both tests.
Ammeter tests will tell you the current in the primary and/or the secondary - this can also help you gain knowledge about the turns ratio. Ammeter tests also tell you that you are at the limit of current for testing with the secondary shorted i.e. at the manufacturer's recommended limit. Don't go higher is the rule.
Voltmeter tests only apply to primary testing when the secondary is shorted and apply to both primary and secondary when the secondary is open circuit.
If you don't have a watt-meter, voltmeter and ammeter tests plus a little bit of common sense and knowledge can help you calculate powers that would be indicated by the watt-meter.
This answer applies to regular power transformers - it isn't intended to cover high frequency transformers although the principles are the same BUT dielectric loss and resonance effects can play a big part.
Best Answer
This can be explained as follows.
Consider the following transformer equivalent circuit (taken from Wikipedia):
The rated power of the transformer, \$P_{n}=V_{1n}I_{1n}=V_{2n}I_{2n}\$, where \$V_{1n}\$, \$I_{1n}\$ are the nominal primary voltage and current, respectively. \$V_{2n}\$, \$I_{2n}\$ are the nominal secondary voltage and current, respectively.
Assuming the high voltage is the primary side and low voltage the secondary side (ie: \$E_{P}>E_{S}\$).
The turns ratio \$(\frac{N_{P}}{N_{S}})\$ is simply the high voltage to low voltage ratio under no load \$(\frac{E_{P}}{E_{S}})\$.
The \$R_{P}\$ and \$X_{P}\$ terms represent the resistance and reactance of the primary coil respectively.
The \$R_{S}^{'}=R_{S}(\frac{N_{P}}{N_{S}})^{2}\$ and \$X_{S}^{'}=X_{S}(\frac{N_{P}}{N_{S}})^{2}\$ terms represent the resistance and reactance of the secondary coil respectively, referred to the primary (ie: they are shown on the primary side of an 'ideal' transformer with turns ratio \$N_{P}:N_{S}\$ by multiplying by a factor of \$(\frac{N_{P}}{N_{S}})^{2}\$).
The \$R_{C}\$ and \$X_{M}\$ terms represent the total core losses (combined eddy current & hysteresis) and magnetisation respectively.
The parameters of the equivalent circuit are determined from an open-circuit test and a short-circuit test.
During the open-circuit test, the rated voltage is applied to the primary side with no load connected to the secondary side. The primary current, voltage and real power into the primary are measured. The current, \$I_{OC}\$ drawn into the primary during the open-circuit test is a fraction (typically 1-6%) of the rated primary current \$I_{1n}\$, so the voltage drop across the primary impedance is negligible. The rated voltage is applied almost entirely across the excitation branch. The real power \$P_{OC}\$ measured during the open-circuit test is therefore very near the core losses (\$I_{C}^{2}R_{C}\$) of the transformer, where:
\$R_{C}=\frac{V_{1n}^{2}}{P_{OC}}\$
So from the open-circuit admittance:
\$Y_{OC}=\frac{I_{OC}}{V_{1n}}\$
We get an estimate for the magnetizing reactance:
\$X_{M}=1/\sqrt{\mid Y_{OC}\mid^{2}-(1/R_{OC})^{2}}\$
During the short-circuit test, a short-circuit link is connected to the secondary winding and a reduced voltage applied to the primary side. A reduced voltage, \$V_{SC}\$ is applied to the primary and adjusted until the rated primary current is drawn. The primary voltage, current and real power \$P_{SC}\$ into the primary are all measured. Since only a fraction of the rated voltage (typically 2-12%) is applied to the transformer, the core-losses are small compared to the copper losses open-circuit test which are large, since the full-load rated current is being drawn. The real power measured during the short-circuit test therefore represents the copper losses (\$I^{2}R\$) in the transformer.
\$R_{eq}=\frac{P_{SC}}{I_{1n}^{2}}\$
The transformer impedance is given by:
\$Z_{eq}=\frac{V_{SC}}{I_{1n}}\$
and the total reactance is:
\$X_{eq}=\sqrt{\mid Z_{eq}\mid^{2}-R_{eq}^{2}}\$
These can be divided equally between primary and referred secondary (\$Z_{P}=Z_{S}^{'}\$), to give:
\$R_{P}=R_{S}(\frac{N_{P}}{N_{S}})^{2}=\frac{R_{eq}}{2}\$
\$X_{P}=R_{S}(\frac{N_{P}}{N_{S}})^{2}=\frac{X_{eq}}{2}\$
For Question 2:
The relative short-circuit voltage \$u_{SC}\$ is just the fraction of the rated primary voltage required to give rated current during the short-circuit test, typically 2-12%.
\$u_{SC}[in \%]=\frac{V_{SC}}{V_{1n}}\times 100\%\$
where \$V_{SC}\$ is the voltage applied during the short-circuit test to get rated current through the primary, \$V_{1n}\$ is the rated (nominal) primary voltage.