Does an electron move from one excitation state to another, or jump? I'm wondering, when an electron changes state, does it move from one state to another over some (very small) time period? Or does it change from one state to another in no time? If the former, what does it mean for it to be in-between states (for however short a period of time)? If the latter, how does it teleport?
(Does this question make sense?)
 A: You already have some very erudite and good answers. I will give an experimentalist's point of view:

I'm wondering, when an electron changes state, does it move from one state to another over some (very small) time period? Or does it change from one state to another in no time?

An electron is par excellence an elementary particle and it is quantum mechanics that reigns here. First of all a bound electron in an energy state is not moving in three dimensional space and time the way a billiard ball moves. When bound it is in orbitals, i.e. it has a probability of being found in a certain $(x,y,z,t)$ when measured, and the measurement will disturb the quantum state. Any time measure will be within the Heisenberg Uncertainty relation $Delta(e)detlat(t)>\hbar$ , i.e. again a probable quantity.


The shapes of the first five atomic orbitals: 1s, 2s, 2px, 2py, and 2pz. The colors show the wave function phase. These are graphs of $ψ(x,y,z)$ functions which depend on the coordinates of one electron. To see the elongated shape of $ψ(x,y,z)^2$ functions that show probability density more directly, see the graphs of d-orbitals below.

So already the concept of "motion" has to be modified for the microcosm of elementary interactions.
In a similar way, if one were to solve for the $(x,y,z,t)$ position of the electron kicked by a photon to another orbital, one would again get a distributions in space and time that would indicate to the experimentalist the probability of finding the electron in that specific $(x,y,z,t)$ were he/she to do an experiment. Probability, not certainty.

If the former, what does it mean for it to be in-between states (for however short a period of time)?

This indeterminacy of an exact position and time has to do with the Heisenberg Uncertainty Principle too,  the position will be within the $σ_x σ_p$ limit given by the HUP.

If the latter, how does it teleport?

It uses the energy given by the incoming photon to rise to the higher energy, but it is a probabilistic superposition of states that does the "motion" which can only be estimated by the HUP limits when either energy or time is measured (and the measurement will drastically change the system).
A: In the case of an electron being perturbed by a sinusoidally varying electric field, you can use perturbation theory to show that it enters a superposition of the two states and oscillates back and forth between the two until it settles down in the final state. This depends on the frequency of the perturbation being equal to the difference in energy levels, as well as things like angular momentum being conserved. Also, in this particular case, the perturbation is time symmetric so both absorption (gain in energy) and stimulated emission (loss of energy) can occur. This is the basis of lasers.
Here's the nitty gritty
Even though the situation is time symmetric, thermodynamics is not, so the probability of emission vs absorption depends on the number of atoms in each state (e.g. Boltzmann statistics in the case of thermal equilibrium or population inversion in the case of lasers.)
A: Leaving aside the quantum measurement problem (i.e. whether or not there is a "collapse" of quantum state to an eigenstate of an observable on measurement) and talking wholly about quantum state between "measurements" and its unitary evolution, I'd say that the transition is definitely a smooth shifting from one "eigenstate" to another, so that the electron's wavefunction is of the form $\alpha_1(t)\, e^{-i\,\omega_0\,t}\,\psi_0(x) + \alpha_1(t)\,e^{-i\,\omega_1\,t}\, \psi_1(x)$ where $|\alpha_0|^2 + |\alpha_1|^2 = 1$, $\alpha_1(0) = 1, \alpha_0(0) = 0$, $\alpha_1(t) \to 0, \alpha_0(t) \to1$ as $t\to\infty$ and $\psi_1, \psi_0$ are the would be "jumped between" "eigenstates" (here I'm thinking of a downward transition from a raised state $\psi_1$ to a ground state $\psi_0$).
In the following I shall stick to the question of an electron as belonging to an atomic or molecular system, rather than bare electron - EM field interaction as in QED. This typifies the kind of system that your question makes sense for i.e. where the electron must have discrete, bound states.
So am using "eigenstates" in quotes because the atom (or molecule - I'll call them all atoms for our purposes) is coupled to the electromagnetic field. So "eigenstate" means, for example, "eigenstate as calculated by the "bare" Dirac equation for an electron in an atomic system sundered from the rest of the Universe. It is no longer an eigenstate of the whole, coupled system, which is why the transition happens.
Lionel's answer gives you a thorough description of how light is absorbed by way of the chapter "Semiclassical theory of light-matter interactions" downloaded through his link to the "Photonics 1" section of the Faculty of Physics, Ludwig Maximilian University, Munchen download section. Here the Fermi Golden Rule is derived for spontaneous absorption rates as well as the time varying coefficients $\alpha_j(t)$ that show you how the transition, although fantastically swift, is nonetheless smooth.
A complementary process, spontaneous emission of a photon from an electron in an excited state also lets you understand this smoothness as well as why the process is one-way. You can look up Wigner-Weisskopf theory for this transition:
V. Weisskopf and E. Wigner, Z. Phys. 63, 54 (1930)
or you can retell this tale through my own simplification presented in J. Opt. Soc. Am. B, Vol. 24, No. 6 June 2007 pp1369-1382. The Weisskopf - Wigner paper unfortunately is in German, which is a shame (for us English speakers) because it is the very best and clearest exposition I know of (as with nigh anything Wigner had a hand in). You can try section  6.3 in chapter 6 of Scully and Zubairy, "Quantum Optics" but this doesn't do it for me: maybe it'll work for you.
So, for now, here is my own summary from JOSA-B.
Let's think of $\hat{a}_1^\dagger$ is thought of as the creation operator that raises the atom in question from its ground state to its first raised state and $\hat{a}_\pm^\dagger(\omega)$ the corresponding operator for a photon in a one dimensional quantized EM field at frequency $\omega$ and in right (+) or left (-) circular polarisation, the Hamiltonian has the form:
$$\hat{H} = \hbar\left(\omega_1 \hat{a}_1^\dagger \hat{a}_1 + \int_0^\infty \omega\,\left(\hat{a}_+^\dagger(\omega) \hat{a}_+(\omega)+ \hat{a}_-^\dagger(\omega) \hat{a}_-(\omega)\right)\,\mathrm{d}\omega +\\ \int_0^\infty \left(\kappa_+(\omega)^*\, \hat{a}_1^\dagger \hat{a}_+(\omega) + \kappa_+(\omega) \hat{a}_+^\dagger(\omega)\,\hat{a}_1\right)\,\mathrm{d}\omega + \int_0^\infty \left(\kappa_-(\omega)^*\, \hat{a}_1^\dagger \hat{a}_-(\omega) + \kappa_-(\omega) \hat{a}_-^\dagger(\omega)\,\hat{a}_1\right)\,\mathrm{d}\omega + const\right)\quad\quad\quad(1)$$
where $\kappa_\pm(\omega)$ is the coupling strength between the excited atom and the free photon electromagnetic modes. The ground state energy for the EM modes is represented by the constant I do not name here. For now, think of this as a coupling to a cavity wherein there is only one electromagnetic mode for each frequency $\omega$. Now I just write this down as a general linear coupled model and suavely make the assertion that the $\kappa_\pm(\omega)$ can be calculated in principle from quantum electrodynamics and thus haughtily give the impression that I know how to do such a thing as a triviality (I don't fully!). With only one photon in the system (i.e. initially in the excited atom and being spontaneously emitted into the field) and given that the above Hamiltonian conserves photon number (adds a photon whenever one is taken from somewhere else), we can reduce the whole system state to the probability amplitude $\psi_1(t)$ of the emitter atom's being excited together with the continuous functions $\psi_\pm(\omega)$ which are the probability amplitudes to find the photon in the mode with frequency $\omega$ and in left and right hand circular polarization, so we don't end up with horrendous complexity explosion wrought by tensor products of electron and photon quantum states:
$$\begin{array}{lcl}
i\,\mathrm{d}_t\, \psi_1(t) &=& \omega_1 \,\psi_1(t) + \int_0^\infty \left(\kappa_+(\omega)^* \,\psi_+(\omega, t)+\kappa_-(\omega)^* \,\psi_-(\omega, t)\right)\, \mathrm{d}\omega\\
i\,\partial_t \,\psi_\pm(\omega, t) &=& \omega\, \psi_\pm(\omega, t) + \kappa_\pm(\omega) \,\psi_1(t)\end{array}\quad\quad\quad(2)$$
You can intuitively see that this equation applies for any number of modes in a quantisation volume, not only a one-mode cavity, because we can absorb the appropriate "degeneracy" co-efficients into the co-efficients $\kappa$ (see my JOSA-B paper if you want to see the details of how this works for a full EM field, but I can assure you it's not exactly riveting stuff!). Now, I show how to solve this system of equations in the section "The Shape of the Spectrum without a Cavity" in this answer. The outcome is:
$$\begin{array}{lcl}
\psi_1(t) &\approx& \exp\left({-i\,\omega_1\,t-\frac{t}{2\tau}}\right)\\
\tau      &\approx& \left.{\frac{1}{2\,\pi\,\left(\kappa_+(\omega)^2+\kappa_-(\omega)^2\right)}}\right|_{\omega=\omega_1-\omega_0}\\
\psi_+(\omega) = \psi_-(\omega) &\approx& \sqrt{\frac{\tau}{\pi}} e^{-i\,\omega\,t}\left(1-e^{-\frac{t}{2 \tau }}\right) \frac{1}{2\tau(\omega -\omega_0)+i}\\
\psi_0(t) &=& e^{-i\,\omega_0\,t+i\theta_0}\,\left(1-e^{-\frac{t}{2 \tau }}\right)
\end{array}\quad\quad\quad(3)$$
and thus we get the exponential, memoryless decay of a spontaneously emitting atom and the implied, Lorentzian lineshape. The last relationship in (3) is the inferred probability amplitude that the atom is in its first raised state and so the electron's state is the following, smoothly varying superposition of ground $\psi_{0,electron}(\vec{r})$ and raised $\psi_{1,electron}(\vec{r})$ "eigenstates":
$$\psi_{electron}(t,x) = \exp\left({-i\,\omega_1\,t-\frac{t}{2\tau}}\right) \psi_{1,electron}(\vec{r}) + e^{-i\,\omega_0\,t+i\theta_0}\,\left(1-e^{-\frac{t}{2 \tau }}\right) \psi_{0,electron}(\vec{r})\qquad(4)$$
Here $\theta_0$ is an undetermined phase factor. Note that the linewidth depends only on the strength of coupling $\kappa_\pm(\omega)$ in the neighbourhood of the uncoupled transition frequency $\omega_1-\omega_0$ defined by the atom's transition energy level difference. It does NOT depend on the shape of the coupling  $\kappa_\pm(\omega)$ as long as this latter is broadband. What's going on intuitively? The atom is coupled to all modes roughly equally. However, it cannot emit into all equally, because if it couples to a frequency away from $\omega_1-\omega_0$, destructive interference hinders the process. So only frequencies near $\omega_1-\omega_0$ are excited. The behavior of Eq. (4) implies a Lorentzian lineshape in the frequency domain, thus we can understand the mechanisms behind the most common spontaneous emission lineshape.
The thermodynamic considerations in Lionel's answer can be readily understood here. Here the raised state is coupled to a continuum of modes. The beginning state, namely with the excitation confined to the atom is a low entropy state (low uncertainty of where the excitation is), and it deforms smoothly and inexorably to the high entropy state wherein the excitation is in a quantum superposition spread over a huge set of electromagnetic field modes. 
A: I would say an electron moves from one state to another over some time period, which is not less than the so called natural line width. If you ask me, in-between states are superpositions of energy eigenstates. I have no idea why those superpositions are less legitimate than energy eigenstates.
A: Very intuitive. No maths. There is an excited state with a symmetrical probability distribution and no e/m dipole moment. There is a ground state (or less excited state) also with no dipole moment. There is a tiny probability that the excited state electron will be in the ground state that allows both states to be present at the same time producing a finite rotating dipole moment which radiates energy and erodes the probability of being in the excited state and augments that of being in the ground state until the ground state probability becomes 1. Thus the photon is emitted with a Gaussian like amplitude profile. If the transition probability is low (e.g. for metastable states) the photon will be slowly emitted with low amplitude, many cycles and a hyperfine spectrum. More probable transitions will produce short fat photons with broader spectral lines.
So it's not the electron that moves between permitted quantum states, it's the probability.
