How do light waves get their size? An atom or (small) molecule has the size of about 100pm. Elektromagnetic waves range from about 0.1nm up to 1 km. The most common way waves (like light) are caused by 'jumping' electrons to another orbit, so their 'jump' is probably smaller than 100pm. 
But how do they succeed in creating waves with a length of 700nm, about a 1000 times larger than the 'jump' they make?
 A: This is essentially the same question as "how does a tiny dipole antenna transmit a wave of several kilometers wavelength"? If you're designing an antenna to transmit a wave, you can indeed get a dipole a few centimeters long to transmit low frequency RF, say less than one MHz. Many commercial transceivers do this; it's not ideal, and you need to drive the antenna with a great deal of current (including it in a resonant circuit helps), but it can certainly be done. The "efficiency" of the process is measured by the antenna's radiation resistance, which measures the work that the antenna can do on the radiating electromagnetic field for a given current. For a short dipole approximation the radiation resistance is
$$R_{rad}=\frac{\pi}{6}\, \mathcal{Z}_0\, \left(\frac{L}{\lambda}\right)^2$$
where $\mathcal{Z}_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}\approx 377{\rm \Omega}$ is the characteristic impedance of freespace. You can see that all nonzero length antennas have a radiation resistance: it's just very small for lengths a great deal less than a wavelength, but, as I said, you can make up for this inefficiency by including the antenna in a resonant circuit.
Likewise, the fluorescing or relaxing excited atom looks, from the EM field point of view, like a tiny dipole antenna. To get more intuition for this idea, I'd urge you to look up the solution of Maxwell's equations for the dipole antenna and study it in detail. Spherical wave solutions certainly exist at all wavelengths for arbitrarily small dipoles: near the source at distances much less than a wavelength, there is a nearfield which is nonpropagating, but the wave becomes more like a plane wave as it propagates over several wavelengths. Many undergraduate physics texts will show you this solution and there are many electromagnetics courses online to help you here.
A: You seem to have some fundamental misunderstandings about the nature of electromagnetic radiation. 
The electron transition describes a change in energy that leads to the emission of a single photon.
The energy of a photon is inversely proportional to the wavelength of the light.
So the light "gets" its wavelength from the energy it carries and not from the distance between quanta.
A: It's like this: 
The speed of light is constant c. Now, during the time when one electron 'jumps' from a higher state $\psi_2$ to lower state $\psi_1$, it oscillates from those states at a frequency equal to $\nu$ in $E = h\nu$, the energy of the photon released. The oscillation induces the formation of electromagnetic waves. So, the frequency of a light wave is actually comes from the frequency of oscillations of the electron. Now the wavelength of a travelling wave is $\lambda = c/\nu$. This is 'not' the diameter of the photon, but rather the length wherein there is one complete cycle of the oscillation of electric and magnetic fields of the light wave.
 
In the case of macroscopic oscillations (radio transmitters), the same frequency of oscillating current in the circuit is needed to produce the same wavelength.
A: Light is an emergent collective phenomenon from zillions of photons. The relation of the energy of the photon to the frequency of the wave is E=h*nu. The photon itself just has spin and energy when measured. Its wavefunction though has a complex dependence that does contain information which will build up the emergent beam with frequency nu.

A detailed QFT explanation can be found here.
A hand waving analogue is the water waves, with meters wavelength even though the molecules move maybe micrometers. (different mathematics but conceptually the same logic)
