As WillO points out in the comments, the quote is actually wrong: non-commuting observables can have (some) simultaneous eigenstates--but a better statement would be that not all of their eigenstates are simultaneous. In other words, if observable $\hat{A}$ has an eigenstate that's a non-trivial superposition of eigenstates of observable $\hat{B}$, then they're non-commuting, and vice versa.
The position-space and momentum-space wavefunctions are Fourier transforms of one another. In position-space representation, a state of definite position $x'$ would be a Dirac delta, $\psi_{x'}(x) = \delta(x-x')$, and the and a state of definite momentum $p'$ would be $$\psi_{p'}(x) = \exp(ip'x/\hbar)/\sqrt{2\pi\hbar}\text{.}$$
It is therefore by default a superposition of position eigenstates.
In fact, when you're writing a wavefunction in position-space, you are automatically representing it as a superposition of position eigenstates, with the function value $\psi(x)$ as the coefficient of the eigenstate of definite position $x$ in that superposition. That's what a wavefunction means: for arbitrary state $|\psi\rangle$, the position-space wavefunction is $\psi(x) = \langle x|\psi\rangle$, where $|x\rangle$ is a position eigenstate.
A similar statement would be correct for momentum-space: momentum eigenstates would be Dirac deltas in momentum-space representation, etc. The only difficultly is that that states like $\psi_p'(x)$ is not normalizable--but that's not a big deal; it's completely dual to the problem of the Dirac delta not actually being a function. They still make sense for the things they're practically used for, as distributions or various other more mathematically sophisticated ways of dealing with that issue.
I've always thought such properties are classical and with continuous range of possibile values.
You seem to be interpreting 'quantum' as implying 'discrete'. This would be very mistaken, and observables in quantum mechanics can and often do have continuous spectra--i.e., a continuous region of allowed results of measurements (eigenvalues). Typically, quantum-mechanical position and momentum observables are continuous. The difference between classical mechanics is that the operators that represent those observables are non-commutative.
This seems to state that the properties obeying the uncertainty principle (position, momentum, etc.) are quantum observables. This would imply that each value we can measure for such property is orthogonal to all others.
It's completely the opposite of this. If you have two observables $\hat{A}$ and $\hat{B}$, with eigenstates (states of definite values) of one that are all orthogonal to eigenstates of the other, then they don't have any non-trivial uncertainty relation between them.
I couldn't find any other information about the relationship between superposition and uncertainty.
One thing that might be helpful to realize is that the observables-as-operators formalism of quantum mechanics in Hilbert space applies just as well to classical mechanics, too--the difference is entirely in what operators correspond to which physical observables, including which ones you are allowed to measure.
For example, if you have an observable $\hat{z}$ with two eigenstates $|z_+\rangle$ and $|z_-\rangle$, an arbitrary superposition is always a valid state, e.g. the following two particular ones
$$|x_\pm\rangle = \frac{1}{\sqrt{2}}\left(|z_+\rangle \pm |z_-\rangle\right)\text{.}$$
Now, in quantum mechanics, you could easily have an operator such as
$$\hat{x} = |z_+\rangle\langle z_+| + |z_-\rangle\langle z_-|$$
that has exactly those two superpositions as its eigenstates. This operator would not commute with $\hat{z}$, and there will be a non-trivial uncertainty relationship between them. The difference is that in standard classical mechanics, something like $\hat{x}$ would not correspond to a valid physical observable. You would simply be forbidden from measuring it.
Instead, if you're only limited to measuring $\hat{z}$ and observables that commute with it, the states $|x_\pm\rangle$ would be indistinguishable from the mixed state of $|z_+\rangle$ and $|z_-\rangle$ with probability $1/2$ each. But quantum mechanics allows you more freedom.
An even easier example: if $|\psi\rangle$ is a non-trivial superposition of multiple eigenstates of some some observable $\hat{A}$, then the projector $\hat\pi = |\psi\rangle\langle\psi|$ is a valid observable in quantum mechanics that doesn't commute with $\hat{A}$ and with a single eigenstate of $|\psi\rangle$.
Best Answer
This shows how the wavefunction, behaves when meeting a barrier. The wave function is the solution of the appropriate quantum mechanical differential equation , with boundary conditions (the barrier).
Note that the energy of the tunneling particle is constant. What varies is the wavefunction whose complex conjugate squared gives the probability of finding the particle at a particular x in this one dimensional example.
Superposition of wave functions, i.e. addition of different wave functions can form interference patterns which, when the summed function is complex congugate squared will show the wave nature of the solutions in interference patterns, i.e. the probability of the particle appearing at x,y,z. This MIT video shows the interference appearing after the superposition of two coherent laser beams.
Here is a definition:
The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for the product of the uncertainties of these two measurements. There is likewise a minimum for the product of the uncertainties of the energy and time.
I hope the above is an adequate description of the Heiseberg uncertainty. Further in the link, it shows the connection with the wavefunction solutions
You state:
The statement in italics is wrong. For tunneling there is no change in "momentum" because there is no change in energy levels. It is based on the Heisneberg uncertainty principle (HUP) only in so far that the uncertainty principle can be derived from the wavefunctions (the second page in the last link). It explains the wavepacket model of a particle.
When the wave packet hits the barrier, a boundary value solution, the top diagram of tunneling holds, and a probability of tunneling will add up from the superposition of each individual wavefunction that mathematically composes the wavepacket .
In conclusion, tunneling depends on probability, not energy and/or momentum, and probability depends on the boundary conditions. The HUP allows it because it is an envelope that can describe the probabilistic particle behavior in general. For details one needs solutions of the appropriate differential equations.