There are two reasons why your earlier question wasn't about radio. The first is, that radio officially goes from 3kHz to 300GHz. The second is, that a transformer is based on a different principle than radio waves. That second reason is what's your question is about: a transformer is based on electromagnetism, radio waves are based on electromagnetic radiation.
Understanding on this topic is really hard, and exists for many people on a lot assumptions. I'll try to give an easy explanation for a layman, for which you'll have to accept some more assumptions than for the detailed explanation below.
Layman explanation
As you know, a magnetic field means that some materials like metals are attracted by others. One can generate a magnetic field by letting an alternating current flow through a wire or coil. That is what happens in the primary coil of a transformer. The other way around, a change in a magnetic field will generate a current in a coil - that's what happens in the secondary coil. These properties of magnetic fields and current are called electromagnetic induction.
Electromagnetic radiation is a particular form of the electromagnetic field. In electromagnetic radiation, the magnetic field will create an electric field (just assume that), but further away from the conductor that began with making the electromagnetic field. The electric field will create a magnetic field, even further away, and so on. It just goes on and on, due to specific properties of the field. That's the key to electromagnetic radiation.
When you are testing with a transformer, the secondary coil exists inside one wavelength of the wave that is produced. This means that the current in the secondary coil does not exist because of electromagnetic radiation, but because of electromagnetic induction: the fields don't create each other.
You can only prove the existence of electromagnetic radiation by transporting waves over more than one wavelength - only then, you can be sure the fields create each other.
Detailed explanation
There is some confusion here, and the cause of that is that the theoretical principle behind radio waves, and radio frequency, don't necessarily go together. Take a look at the Radio Wikipedia:
Radio is the wireless transmission of signals through free space by electromagnetic radiation of a frequency significantly below that of visible light, in the radio frequency range, from about 30 kHz to 300 GHz. These waves are called radio waves. Electromagnetic radiation travels by means of oscillating electromagnetic fields that pass through the air and the vacuum of space.
Note: I believe the 30kHz minimum should be 3kHz (reference: here and here)
You can see that there might be other waves, based on the same principle and working the same way, with a frequency <3kHz or >300GHz, that are just therefore not part of "Radio". Those waves aren't radio waves and they aren't in the RF spectrum, but they are just the same, when you forget about the frequency.
But there's more! Radio waves are electromagnetic radiation. Electromagnetic radiation contains of two components, one electrical and one magnetic. These components create each other, as said above. The red magnetic field creates a blue electric field, which creates the next magnetic field, and so on.
From the Electromagnetic radiation Wikipedia:
Electromagnetic radiation is a particular form of the more general electromagnetic field (EM field), which is produced by moving charges. Electromagnetic radiation is associated with EM fields that are far enough away from the moving charges that produced them that absorption of the EM radiation no longer affects the behaviour of these moving charges.
What we were trying to do in your earlier question was really just picking up the weak magnetic field, because that's what a secondary coil does.
I guess you're now wondering: but does a transformer do electromagnetic radiation, or is it just a magnetic field? Let's have a look, with the Electromagnetic radiation Wikipedia:
... the electric and magnetic fields in EMR1 exist in a constant ratio of strengths to each other, and also to be found in phase ...
1: electromagnetic radiation, compared to the electromagnetic field - note by author
Think about the transformer. A magnetic field is generated when the current changes. Let's say we have a pure sine as the current, \$I(t)=sin(t)\cdot{}c\$. We can get the change of the current on a specific moment by taking the derivative of that sinus, which is the cosine, so: \$B(t)=cos(t)\cdot{}c\$. Now have a look at the functions \$I(t)\$ and \$B(t)\$, which should exist in "a constant ratio of strengths to each other" and in phase.
Note: the constant \$c\$ is because the formulas depend on other things as well, that are irrelevant now and constant in a specific situation
You can already see those functions aren't in phase. They aren't in a constant ratio to each other either. You can see that by plotting \$f(t)=\frac{sin(t)}{cos(t)}=tan(t)\$:
So no, a transformer does not radiate electromagnetic radiation. The waves aren't in a constant ratio of strength to each other, neither are they in phase. The tests you did with a transformer in your earlier question, were just based on a magnetic field.
This difference between picking up a magnetic field and magnetic radiation is known as the difference between near and far field.
Summary
There are two main reasons why your experiments weren't about radio. The first is that it just was the wrong frequency. The second is that a coil with an AC current does not provide electromagnetic radiation.
Reference
H is the driving force in coils and is ampere turns per metre where the metre part is the length of the magnetic circuit. In a transformer it's easy to determine this length because 99% of the flux is contained in the core. A coil with an air core is difficult as you might imagine.
I think of B as a by-product of H and B is made bigger by the permeability of the core.
In electrostatics, E (electric field strength) is the equivalent of H (magnetic field strength) and it's somewhat easier to visualize. Its units are volts per metre and also gives rise to another quantity, electric flux density (D) when multiplied by the permittivity of the material in which it exists: -
\$\dfrac{B}{H} = \mu_0\mu_R\$ and
\$\dfrac{D}{E} = \epsilon_0\epsilon_R\$
Regarding ferrite data sheets, the BH curve is the important one - it tells you the permeability of the material and this directly relates to how much inductance you can get for one turn of wire.
It will also indicate how much energy could be lost when reversing the magnetic field - this of course will always happen when ac driven - not all the domains in the ferrite return to produce an average of zero magnetism when the current is removed and when reversing the current the remaining domains need to be neutralized before the core magnetism goes negative - this requires a small amount of energy on most ferrites and gives rise to the term hysteresis loss.
Other important graphs in a ferrite data sheet are the permeability versus frequency graph and permeability versus temperature.
From personal experience of having designed a few transformers, I find them tortuous in that I never seem to naturally remember anything other than the basics each time I begin a new design and this is annoying - in this answer I had to double check everything except the units of H!
Best Answer
The ratio of mutual flux to primary flux is the same as the coefficient of coupling. However, the flux linkage of the secondary is the mutual flux multiplied by the number of turns of the secondary. Therefore the flux linkage ratio is not the same as the coefficient of coupling and may have a value greater than 1 unlike the coefficient of coupling which is always less than 1 and greater than 0. Lets prove that out from some known laws.
So from Faraday's law we can derive the following relationship between the mutual inductance and the mutual flux:
\$ L_m = N_s \cdot \frac{\delta \phi_m}{\delta i_p} \$
the mutual flux is the portion of the flux from the primary that overlaps with the secondary. Therefore the following is always true:
\$ \phi_m \leq \phi_p\$
We also know flux as a function of current can be derived from the following equation:
\$ L = \frac{\phi \cdot N}{i} \$
Solving for the flux we get:
\$ \phi = \frac{L \cdot i}{N}\$
Lets also define the ratio of flux linkage with the secondary coil to primary coils flux:
\$ R_\lambda = \frac{\lambda_s}{\phi_p} \$
Solving for the flux linkage and substituting the primary flux with the earlier equation we get:
\$ \lambda_s = R_\lambda \cdot \phi_p \$
\$ \lambda_s = R_\lambda \cdot \frac{L_p \cdot i_p}{N_p} \$
\$ \lambda_s = \frac{R_\lambda \cdot L_p}{N_p} \cdot i_p \$
It is important to note here that the flux linkage, \$\lambda_s\$, is the mutual flux multiplied by the number of turns of the secondary winding:
\$\lambda_s = N_s \cdot \phi_m\$
Alternatively:
\$\phi_m = \frac{\lambda_s}{N_s}\$
Since we know that flux linkage of the secondary is the portion of the primaries flux which is mutual multiplied by the number of turns in the secondary, then it stands to reason the ratio between the flux linkage and the primary flux can be any positive value, including values greater than 1.
Next we can take the derivative of our first equation by substitution:
\$ L_m = N_s \cdot \frac{\delta}{\delta i_p} (\frac{\lambda_s}{N_s}) \$
\$ L_m = N_s \cdot \frac{\delta}{\delta i_p} (\frac{\frac{R_\lambda \cdot L_p}{N_p} \cdot i_p}{N_s}) \$
\$ L_m = N_s \cdot \frac{\delta}{\delta i_p} (\frac{R_\lambda \cdot L_p}{N_p \cdot N_s} \cdot i_p) \$
\$ L_m = N_s \cdot \frac{R_\lambda \cdot L_p}{N_p \cdot N_s} \$
\$ L_m = \frac{R_\lambda \cdot L_p}{N_p} \$
We also know the relationship between mutual inductance and the coefficient of coupling with this equation:
\$ L_m = k \cdot \sqrt{L_p \cdot L_s} \$
We can now substitute this in to get an equation showing the relationship between the flux link ratio and the coefficient of coupling:
\$ k \cdot \sqrt{L_p \cdot L_s} = \frac{R_\lambda \cdot L_p}{N_p} \$
Now we can solve for the flux linking ratio and simplify:
\$ k \cdot \sqrt{L_p \cdot L_s} = R_\lambda \cdot \frac{L_p}{N_p} \$
\$ k \cdot \sqrt{L_p \cdot L_s} \cdot \frac{N_p}{L_p} = R_\lambda\$
\$ k \cdot \frac{N_p \cdot \sqrt{L_p \cdot L_s}}{L_p} = R_\lambda\$
Since all of the constants here are positive values we can further reduce this to:
\$ k \cdot N_p \cdot \sqrt{\frac{L_s}{L_p}} = R_\lambda\$
It should now be clear that the flux link percentage (\$R_\lambda\$) is not the same value as the Coefficient of Coupling (\$k\$).
While most of the derivations above I did myself you can find most of the equations I pulled in as laws (Like Faraday's Law in the first equation) in this wonderful online book that is worth taking a read for a deeper explanation of the subject.