If you have only the power rating, you are stuck.
If you have the resistance in ohms as well, you have enough information via either of the equations:
$$P = \frac{V^2}{R}$$
or
$$P = I^2 R$$
So given a 2 Watt 8 ohm speaker,
$$V^2 = P \cdot R = 16$$
so V=4 volts, and I = V/R = 0.5 amps
Now for a speaker you also need to know if that was the peak power rating or the RMS (roughly speaking, average) power. If it is 2W "rms" that means 4 volts rms, or 2.8*4 = 11.2 volts peak-peak, which suggests an amplifier running off 12V DC.
Surely the transformer's magnetics couldn't handle these spikes as I'd
expect the core would saturate.
Core saturation has nothing to do with load VA rating. It has everything to do with the magnetization current flowing in the primary. This current is largely constant irrespective of secondary load current.
In short, the ampere turns on the secondary winding (caused by the load) are exactly equal (but opposite in sign) to the ampere turns on the primary due to that secondary load current. Neither of these currents are the magnetization current that can saturate the core.
Imagine a simplified core with a single turn primary: -
At the moment it's just a single turn inductor. With V applied, Im flows and inductance, frequency and voltage all determine how much current (Im) flows. OK so far?
Now imagine that single turn were replaced by 2 closely coupled parallel turns like this: -
You would find that Im/2 flows in each or, in other word,s the same overall current flows. A nice side effect of this is that each individual coil must have twice the inductance of the single coil and, if you happened to make a two turn inductor this way (by wiring them in series) it would have 4x the inductance. Just think about it for a while.
Next scenario: -
So, you drive one of those closely coupled coils and look at the voltage on the other coil. The driving voltage and the secondary voltage are in phase and of equal amplitude (1:1 turns ratio). Do you see why? If not, consider what would have happened in the 2nd scenario if (say) the voltages were out of phase - you'd get a fire and you wouldn't get the inductance rising with turns squared - you'd get zero inductance. This doesn't happen.
Final scenario: -
You've applied a load to that 2nd winding and because in the 3rd scenario you (hopefully) recognized that the voltages were in phase, you have to admit that the currents are COMPLETELY antiphase.
From here, it's a minor leap of faith to recognize that the ampere.turns on the primary (due to the secondary load) are equal and opposite to the ampere.turns on the secondary. As I said earlier, neither of these currents are the magnetization current that can saturate the core - this is due to Im.
It's magnetic field strength that drives the magnetism. It's called "H" and H is measured in ampere.turns per metre. The "per metre" part is irrelevant because it's a core physical dimension and applies equally to primary and secondary.
Basically H never alters one bit due to loading effect. In fact that's not quite true; it gets lower with more load because the copper losses lower the actual terminal voltage and reduce the magnetization current a little bit.
Best Answer
There are many unknown factors. Both power and current will vary with load and there is only one way to find out how they relate: measure them.
One of the most important properties of AC power you neglected is the power factor cos(φ) with inductive loads:
\$P = U \cdot I \cdot \cos(\varphi)\$
\$\begin{align} \cos(\varphi) & = \dfrac{P}{U \cdot I} \\ & = \dfrac{1300\text{W}}{120\text{V} \cdot 12\text{A}} \\ & \approx \boxed{0.9} \end{align}\$
which sounds about what I'd expect for a vacuum cleaner.
Check this Wikipedia article on electric power for more background details.