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.
There are two answers to this question. The first is that, yes, in non-relativistic quantum mechanics, you can have things going faster than the speed of light, because relativity is never taken into account. The fix is to learn quantum field theory.
We can also consider this particular situation more closely. Your thought experiment suggests that a momentum measurement can "teleport" a particle infinitely far away in an infinitely short time, which feels unphysical, regardless of relativity.
The solution is that a precise momentum measurement takes a finite amount of time. (And an infinitely precise momentum measurement, as you're suggesting, takes infinitely long.)
To see this, start from the energy-time uncertainty principle
$$\Delta E \Delta t \geq \hbar$$
where $\Delta E = \Delta (p^2/2m) = p \Delta p / m$. Then we have the bound
$$\frac{p \Delta p \Delta t}{m} \geq \hbar$$
where $p$ is the (average) value of momentum you get, $\Delta p$ is the uncertainty on that momentum, and $\Delta t$ is the time it took to perform the measurement. This tells us that more precise momentum measurements take longer.
Now, the final state after this 'smeared' momentum measurement is a wavepacket centered on the origin with width $\Delta x$, with
$$ \Delta x \Delta p \sim \hbar.$$
Finally, combining this with our other result gives
$$ \frac{\Delta x}{\Delta t} \leq \frac{p}{m}.$$
That is, the particle is not moving any faster than it would be, semiclassically.
Best Answer
$\newcommand{\ket}[1]{\lvert #1 \rangle}$There is no such thing as "looking like a collapsed wavefunction", even if you believe there's collapse.
Let's go to the finite dimensional case and have a simple two-level spin system, that is, our Hilbert space is spanned by, e.g., the definite spin states in the $z$-direction $\ket{\uparrow_z}$,$\ket{\downarrow_z}$.
Now, the state $\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{\uparrow_z} + \ket{\downarrow_z})$ is not an eigenstate of $S_z$, and measurement of $S_z$ will collapse it with equal probability into $\ket{\uparrow_z}$ or $\ket{\downarrow_z}$. Yet, this is an eigenstate of the spin in $y$-direction, i.e. $\ket{\psi} = \ket{\uparrow_y}$ (or down, we'd have to check that by computation, but it doesn't matter for this argument). So, although you can "collapse" $\ket{\psi}$ into other states, it already looks like a collapsed state by your logic. This shows that the notion of "looking like a collapsed state" is not very useful to begin with.
Furthermore, you seem to be confused about the difference between a measurement (inducing collapse in some dictions) and the application of an operator. You say
but this is non-sensical. The action of the Hamiltonian is an infinitesimal time step, as the Schrödinger equation tells you:
$$ \mathrm{i}\hbar\partial_t \ket{\psi} = H\ket{\psi} $$
and, for an eigenstate $\ket{\psi_n}$, which is a solution to the time-independent equation with energy $E_n$, you have by definition $H\ket{\psi_n} = E_n\ket{\psi_n}$, that is, the Hamiltonian is a "do nothing" operation on eigenstates since multiplication by a number does not change the quantum state. That is, after all, why we are interested in the solutions to the time-independent equation - because these are the stationary states that do not evolve in time. This has nothing to do with collapse, or measurement.
Lastly, exactly determinate states of position are not, strictly speaking, quantum states, since the "eigenfunctions" of the position operator "multiplication by x" are Dirac deltas $\psi(x) = \delta(x-x_0)$, which are not proper square-integrable functions $L^2(\mathbb{R})$ as quantum states are usually required to be. But yes, this is "a spike", and conversely, the determinate momentum states are plane waves $\psi(x) = \mathrm{e}^{\frac{\mathrm{i}}{\hbar}px}$.