I think the main issue here is that you're attempting to think about a system in isolation -- the energy states of an electron subject to the electrostatic potential created by the positively charged nucleus -- and trying to understand the measurement based on this system. This is hopeless, as the measurement is not a part of this system.
Your first big red flag should be that energy is not conserved in such a transition. Where does the energy in the difference between the two states go?
You might try thinking about how this measurement is performed in a laboratory. Somewhat tangentially, this was an experiment in my undergraduate physics laboratory, because it encouraged you to think about precisely this type of problem[1].
I should also specify that the Rydberg-Ritz "difference of frequencies associated with the spectral lines" really just means "transitions between two excited states of the atom."
When an electron 'relaxes' from one energy state to another, the difference in energy has to go somewhere (by conservation of energy). As an interesting corollary, such a transition is forbidden if there is nowhere for that energy to go[2]. In this case, the electron's potential energy is released in the form of an excitation of the electromagnetic field, also known as a photon.
On the other end of the experiment, we measure the wavelength of this photon, typically with a grating monochrometer or similar apparatus. Using the energy-wavelength relation for a photon in free space, $E=\frac{hc}{\lambda}$, we calculate the energy of the emitted photons.
But we still have not gotten to the measurement in any true quantum mechanical sense. How do we know which wavelengths (energies) correspond to the transition lines of the atom? We plot the expected number of photons per unit time versus energy and look for spikes that look like a Lorentzian (or really, a Voigt profile). The center of those spikes is the energy we associate with the transition.
So the true measurement we are making is the expected value of the number operator, $\left<\hat{N}\right> = \left< a^\dagger a \right>$ when the monochrometer is set to different wavelength values.
In summary, you are correct that the difference in energy levels does not correspond nicely to a measurement. What does correspond to a measurement is the energy of the emitted photon when an electron traverses that energy difference. By conservation of energy, these must have the same value[3].
As an end note, including the measurement apparatus can be a powerful tool in the analysis of quantum systems. In the field of quantum information, it's typically referred to as the 'ancilla' system and allows you to understand the measurement of POVMs
[1] Technically, we were looking at the effects of the change in nuclear mass between Hydrogen and Deuterium, but that is really a tangent.
[2] This is the basic idea behind suppression of spontaneous emission in the Purcell Effect.
[3] I left out an important bit here -- why the entire energy of the transition must be conveyed to a single photon. This is essentially a consequence of the quantization of energy levels and the linearity of the electromagnetic field, though the ability to explain this simply and accurately lies beyond my skills.
According to Bohr model, the absorption and emission lines should be infinitely narrow, because there is only one discrete value for the energy.
There are few mechanism on broadening the line width - natural line width, Lorentz pressure broadening, Doppler broadening, Stark and Zeeman broadening etc.
Only the first one isn't described in Bohr theory - it's clearly a quantum effect, this is a direct consequence of the time-energy uncertainty principle:
$$\Delta E\Delta t \ge \frac{\hbar}{2}$$
where the $\Delta E$ is the energy difference, and $\Delta t$ is the decay time of this state.
Most excited states have lifetimes of $10^{-8}-10^{-10}\mathrm{s}$, so the uncertainty in the energy sligthly broadens the spectral line for an order about $10^{-4}Å$.
Best Answer
Good question! The function $\psi$ does not need to be Hamiltonian eigenfunction. Whatever the initial $\psi$ and whatever the method used to find future $\psi(t)$, the time-dependent Schroedinger equation $$ \partial_t \psi = \frac{1}{i\hbar}\hat{H}\psi $$ implies that the atom will radiate EM waves with spectrum sharply peaked at the frequencies given by the famous formula $$ \omega_{mn} = \frac{E_m-E_n}{\hbar}, $$
where $E_m$ are eigenvalues of the Hamiltonian $\hat{H}$ of the atom.
Here is why. The radiation frequency is given by the frequency of oscillation of the expected average electric moment of the atom
$$ \boldsymbol{\mu}(t) = \int\psi^*(\mathbf r,t) q\mathbf r\psi(\mathbf r,t) d^3\mathbf r $$ The time evolution of $\psi(\mathbf r,t)$ is determined by the Hamiltonian $\hat{H}$. The most simple way to find approximate value of $\boldsymbol{\mu}(t)$ is to expand $\psi$ into eigenfunctions of $\hat{H}$ which depend on time as $e^{-i\frac{E_n t}{\hbar}}$. There will be many terms. Some are products of an eigenfunction with itself and contribution of these vanishes. Some are products of two different eigenfunctions. These latter terms depend on time as $e^{-i\frac{E_n-E_m}{\hbar}}$ and make $\boldsymbol{\mu}$ oscillate at the frequency $(E_m-E_n)/\hbar$. Schroedinger explained the Ritz combination principle this way, without any quantum jumps or discrete allowed states; $\psi$ changes continuously in time. Imperfection of this theory is that the function oscillates indefinitely and is not damped down; in other words, this theory does not account for spontaneous emission.