The dispersion of a wave is a result of the relationship between its frequency and its wavelength, which is appropriately known as the dispersion relation for the wave. For classical waves this depends on the medium: light, for example, will be dispersionless in vacuum and will have dispersion inside material media because the medium affects the dispersion relation. Quantum mechanical waves, on the other hand, have dispersion fundamentally built in.
Let's have an equation look at how light behaves. The dispersion relation for light is
$$\omega=\frac c n k,$$
where $k=2\pi/\lambda$ is the wavenumber, $c$ is the speed of light in vacuum, and $n=\sqrt{\varepsilon_r\mu_r}$ is the medium's refractive index. In vacuum, $n\equiv1$ and there is no dispersion: the phase and group velocities, $\frac \omega k$ and $\frac{d\omega}{dk}$ are equal, constant, and independent of $k$, which are the mathematical conditions for dispersionless waves. In material media, though, $n$ will depend on the wavelength - it has to depend on the wavelength - and there will be dispersion.
Matter waves, on the other hand, are quite different. What are their frequency and wavelength, anyway? Well, the first is given by Planck's postulate that $E=h\nu$, and the second by de Broglie's relation $p=h/\lambda$; both should really be phrased as
$$E=\hbar \omega\text{ and }p=\hbar k.$$
How are $\omega$ and $k$ related? The same way that $E$ and $p$ are: for nonrelativistic mechanics, as $E=\frac{p^2}{2m}$. Thus the dispersion relation for matter waves in free space reads
$$\hbar\omega=\frac{\hbar^2k^2}{2m},\text{ or }\omega=\frac \hbar{2m}k^2.$$
Note how different this is to the one above! The phase velocity is now $v_\phi=\frac\omega k=\frac{\hbar k}{2m}$, and it is different for different wavelengths.
Now, why exactly does that imply dispersion?
Let's first look at the phase velocity, which is the velocity of the wavefronts, which are the planes that have constant phase. Since the phase goes as $e^{i(kx-\omega t)}$, the phase velocity is $\omega/k$. This, of course, is for a single plane wave, and doesn't apply to a general wavepacket for which phase is not as well defined, and for which the different wavefronts might be doing different speeds.
How then do we deal with wavepackets? The approach that works best with the formalism above is to think of a wavepacket $\psi(x,t)$ as a superposition of different plane waves $e^{i(kx-\omega t)}$, each with its own weight $\tilde\psi(k)$:
$$\psi(x,0)=\int_{-\infty}^\infty \frac{dk}{\sqrt{2\pi}} \tilde \psi(k)e^{ikx}.\tag{1}$$
Now, if all the different plane-wave components $\tilde \psi(k)e^{ikx}$ moved at the same speed then their sum would just move at that speed and would not change shape.
(More mathematically: if $v_\phi=\omega/k$ is constant, then
$$\psi(x,t)=\int_{-\infty}^\infty \frac{dk}{\sqrt{2\pi}} \tilde \psi(k)e^{i(kx-\omega t)}
=\int_{-\infty}^\infty \frac{dk}{\sqrt{2\pi}} \tilde \psi(k)e^{ik(x-v_\phi t)}
=\psi(x-v_\phi t,0),$$
so the functional form is preserved.)
For a matter wave in free space, however, the different phase speeds are not the same, and the different plane-wave components move at different speed. It is the interference of all these different components that makes them sum to $\psi(x,0)$ in equation (1), and if you mess with the relative phases you will get a different sum. Thus, with longer waves moving slower and shorter ones going faster, wavepackets with lots of detail encoded in long high-$k$ tails of their Fourier transform will change shape very fast.
In general it is hard to predict what the evolution of a wavepacket will do to it in detail. However, it is very clear that all wavepackets will (eventually) spread, since some components are going faster than others. Since the total probability is conserved, this must mean that the probability density will in general decrease.
If I put a particle with zero net momentum localized in some interval, then the probability of it remaining there will decrease. Note, though, that this is no surprise! The Uncertainty Principle demands that there be uncertainty in the particle's momentum. There is then some chance that the particle was moving to the left or to the right, so who's surprised to eventually find it out of the original interval?
When the two waves collide, why do they pass right through each other?
The problem in understanding waves, in my opinion, lies in the fact that one often applies the same concepts he uses in describing particles, to waves. Waves are not particles, and particles are not waves.
While this seems a stupid tautology, it's not that easy to stop mixing the concepts and start thinking in the right framework.
Mathematically it's due to the principle of superposition: the sum of the two solutions of a wave equation is also a solution.
Superposition principle is way more fundamental than you could think. It doesn't just tell you that the sum of two solutions is a solution; it tells you that you can think about each wave independently from the other waves, as if the weren't there. You can picture each wave travelling down the wire, and then sum all the waves that compose your whole waveform.
But intuitively it's not clear why the waves would not, say, just cancel each other during the collision.
Start reasoning in terms of waves. How could a single wave be stopped? Only by dissipation in the medium, or by external forces. Not by means of other waves. That's where the superposition principle plays its role. Different waves (in a linear medium) no not interact. That's it. In your example you see a sort of interaction, but it's actually just a coincidence. It's just visual. You are interpreting the waves as interacting, but actually they are ignoring each other and keeping their behavior unchanged. You could test my statement analyzing the two waves in terms of their momentum/wave-vector instead of their "position".
What would be a convincing 'local' explanation - in terms of the individual particles in the medium (or segments of the medium), that move only due to the interactions with their neighbors?
In waves framework a local explanation would be that the effect of each force acting on a particle is independent of other forces acting on the same particle (or other particles in general). In your example this explains precisely why the center particle doesn't move: it is experiencing equal opposite forces on itself, one coming from the right wave and one coming from the other.
One final remark: the requisite that the effect of each force acting on a particle is independent of other forces acting on the same particle is precisely the superposition principle. It's not just a global property, it's a local one. It must hold in each point of the medium in order to hold globally.
I hope this animation helps you to visualize the importance of superposition.
The top plot shows a wave travelling to the right, the middle one shows a wave travelling to the left, identical but of opposed sign, while the lower figure shows the sum of the two waves.
Link to gif animation
The same but with different amplitudes
Link to second gif animation
Spring-mass model
Directly from wikipedia:
The wave equation in the one-dimensional case can be derived from Hooke's Law in the following way: Imagine an array of little weights of mass m interconnected with massless springs of length h . The springs have a spring constant of k:
Here the dependent variable u(x) measures the distance from the equilibrium of the mass situated at x, so that u(x) essentially measures the magnitude of a disturbance (i.e. stress) that is traveling in an elastic material. The forces exerted on the mass m at the location x+h are:
This could be the key point: if you look at $F_{Newton}$ as the effect on the central particle caused by waves passing by, you see need to attribute the cause to $F_{Hooke}$, the elastic force. This effect is precisely linear: how do you tell if the resulting force is caused by a single wave of a certain amplitude, or two waves with different amplitudes that sum to the same amount, or infinite waves that again sum to that total force. You simply can't. There's an infinite number of ways to cause that exact amount of force on the central particle.
Final edit: from the animation it's actually not that clear why the wave shouldn't disappear. It is because you are just looking at the deformation. But it doesn't hold all the information: it's a system evolving in time. You have to also look at speed and force at any instant. This final animation should evaporate all your dilemmas: 
You see that when the waves encounter, speeds and forces add, not elide. The elision you see in the deformation domain is just a "coincidence".
Best Answer
I think your confusion arises because you think that the electron is modelled by the wavepacket. Instead the wavepacket of an electron which we suspect near a point $x_0$ represents (or is associated with) the electron and tells us how likely it is to find the electron near that point $x_0$. This is an important distinction because this way the concept of dispersion makes more sense, since it is nonsensical to say that the electron smears out while it's perfectly valid to say that the wavepacket smears out. So again: the wavepacket is not the electron but represents (is associated with) the electron.
This also means that the wavepacket isn't spherically symmetric just because the electron is "round". Sure it makes sense that we use gaussian packets since for a free particle the probability to find it is equal in all directions (if we start from a point where the probability is a maximum). The point I'm trying to make is that even if you had a "quantum elephant" the wavepacket need not look like an elephant.
4.) I think this is nitpicking, but the correct way to say it should be "...hardness stands for the amplitude of the probability density.", since wavepackets also are of the form $\psi(x,t)=\dots$, where $\left|\psi\right(x,t)|^2=\dots$ is the probability density. So for your hairy ball this means that at the outer layers, where the ball is soft, the probability density is small and the electron unlikely to be found.
5.) This is true, since the coefficient function of a gaussian packet for example is
$$\phi(\vec k)=A^3e^{-(\vec k - \vec k_0)^2 d^2 } ,$$
so while most of the k's are near $k_0$, there are also many k's slightly different from it, although the number of k's much different than $k_0$ decreases exponentially. You can also see it this way: $\vec k - \vec k_0$ is the vector which connects the head of $\vec k$ to the head of $\vec k_0$, when this arrow is big, his square is also big and the coefficient function small.
A.) When you want to stay in your hairy ball analogy, then it would best to think about this ball getting bigger and more soft. So when you have your ball at the beginning and draw a sphere around the center where the ball is hard, then this sphere you draw gets bigger over time and the ball inside the sphere becomes more soft. When you wait long enough the ball becomes very large and very soft everywhere and the center is hardly any harder than the outer layers. If you view it this way then you can call it dispersion.
Below is an image how dispersion can be depicted. The black circle would then be the sphere in your ball while the probability amplitude would correspond to the hardness. The circle (sphere) becomes bigger while the amplitude (hardness) becomes less over time.
The next thing is that the dispersion is proportional to the time elapsed, so even if the wavepacket doesn't move initially the wavepacket spreads out. And no, it's not the electron which explodes: it's the wavepacket which "explodes" in the sense that it spreads out in space. This means that a free particle becomes delocalized: since the wavepacket spreads out so does the probability of finding the electron at some place.
B.) When you form a wavepacket you have to do the following integral
$$\int A(k) e^{i(kx-w(k)t)}d^3 k. $$
Since you want to integrate over all $k$'s, you have to write omega as a function of $k$. Now, since for a wavepacket you need $k$'s very close to some value $k_0$ you can expand the $\omega(k)$ function into
$$\omega = \omega_0 + \left( \frac{d\omega(k)}{dk}\right)_{k_0}(k-k_0)^2+... $$
This is why the wavepacket has only one single frequency $\omega_0$ (first term): you assume the values of $k$ are centered close to some $k_0$, so $\omega(k)$ is also centered about some $\omega_0$, while the distribution of $k$'s around $\omega_0$ - the dispersion relation - gives you the velocity at which the packet moves (second term).