What exactly is reactive power, in concrete terms?

The imaginary part is energy that is bouncing back and forth between reactive components without being dissipated. Essentially, you can think of it as energy that gets stuck in the wiring due to the existence of capacitance and inductance.

Now to your actual question:

I am trying to understand how the mathematical expression Imaginary(s) is what you just described vaguely.

Complex power = V^2 / impedance. The impedance is a complex value, with the real part being resistance. The imaginary part is reactance. If you take the imaginary part of complex power, you get V^2 / imag(impedance) = V^2 / reactance. So the imaginary power is just the effect of voltage on reactance (the imaginary part of impedance).

I quess some images can be useful. The wanted formula is at the end.

V1 is 100V peak sinusoidal voltage f=50Hz. There's 100 Ohm resistor R1 and 0.5H inductor L1 as load. The load current is about 530mA peak and its phase lags from V1 as it should be with inductive load.

In the next image the red curve is the momentary power from the source to the load i.e. the voltage of node 1 multiplied by the current of R1:

We see that the power is flowing in 50% shorter cycles and it is a part of the time negative. Negative power means that the load sends energy back to the source. That energy was stored in the magnetic field but it was not dissipated.

Most of us can surely believe that reactive power is the power which goes from source to the inductor and back. But without a formula that's only qualitative definition. It's wrong to say it's the negative part because there's all the time some dissipation in R1, too. The red curve is only the net flow from the source, one cannot easily see from it the reactive power. The real power can be seen easily. It must be the average of the curve.

Real power and reactive power can be calculated with the phasors but they can be also drawn in the simulator:

The upper curve is the dissipated power in R1, it's the voltage over R1 multiplied by the current. The peak dissipated power is 28.5 Watts.

The lower curve is the power into L1. It's half of the time negative because L1 returns all it has received. The peak value is in this case about 22.5 watts.

But how the red curve is related with the usual average reactive power which can be calculated with RMS phasors as Im(S)? The result is surely somehow surprising. The peak of the red is equal with the reactive power.

The lack of any numeric multiplier can be proven with trigonometric formulas. The red curve is the product of the current and the voltage of L1. Those quantities have 90 degrees phase difference. If we discard the common part of their phase angles we can see that the red curve is (Up)(Ip)cos(at)sin(at) where Up and Ip are the peak voltage and peak current and a is 2Pi*frequency.

cos(at)sin(at) is equal with sin(2at)/2 so the red curve has peak value UpIp/2. But that's the product of RMS values of inductor's voltage and current and that's the reactive power.

As a conclusion we can formulate the next rule: The absolute value of the reactive power Q=Im(S) (sign omitted) is physically the peak value of the total energy flow (=watts) from the source to the fields of the reactive components of the load when flow to inductances is considered to be positive and flow to capacitors is considered to be negative.

One should note that the load is handled as one. The input reactive power contains zero info of what happens inside the load. There can for ex. be resonant circuits where substantial energies oscillate between inductors and capacitors, but the total input reactive power can still be zero.

Reactive power is when X amount of REAL power flows to the load, does no work, and then X amount of REAL power flows back to the source. The reactive power in that case is X.

That means that in order to view reactive power you have to observe over some time span, since if you observe a sufficiently short time span all you will see is real power flowing one way or the other and miss out on its return trip.

Mathematically, for any arbitrary voltage and current waveforms (not just sine waves) you can construct the power waveform by calculating \$V \times I\$ for all instants in time using the instantaneous voltage and current at from moment to moment. Each one of those results is real power at that instant in time flowing one way or the other.

I am going to define positive VI values as that which flows from source to load, and negative VI values as that which flows from load to source. If you took all the negative VI values and are able to find enough positive VI to cancel it out with, then the sum of that negative VI (or the sum of the positive VI with which it cancelled with) is reactive power.

If you could NOT find enough positive VI to completely cancel out the negative VI, then guess what? You actually have more power flowing from your load to your source than the other way around which means that you got your load and sources reversed. The negative and positive portions that were able to cancel out are still reactive power.

The average of the instantaneous VI ends up cancelling out the negative and positive instantaneous VI values leaving you with some net amount. This net amount is the power that flowed from the source to the load and then never returned (because it was dissipated in the load since it did work). Therefore, the average of instantaneous power is real power flowing from the source to the load, since averaging cancels out and ignores the power that flows back and forth does no work

RMS voltage and currents lose this phase information so when you multiply RMS voltages and currents you get the maximum possible power (i.e. if the voltage and current were in phase). The so-called "apparent power".

So you could think of reactive power as a book-keeping method for the flow of real power that does no work due to redundantly flowing back and forth. So reactive power isn't imaginary at all. It's real power that is just spinning its wheels.