I think the wave equation can by derived from geometry alone, without using physics. Consider $f(x-ct)$ and consider small changes in $x$ and $t$, ie. $\Delta x$, $\Delta t$ (They each cause a small shift or translation of $f(x-ct)$). Note that $\Delta x$ = $c\Delta t$. So $\frac{\Delta f}{\Delta x}$ = $\frac{\Delta f}{c\Delta t}$ = $\frac{1}{c}\frac{\Delta f}{\Delta t}$. Doing that again we get
$$\frac{\Delta^2 f}{\Delta^2 x} = \left(\frac{1}{c}\right)^2\frac{\Delta^2 f}{\Delta^2 t}$$. Then letting $\Delta$ become very small we get
$$
\frac{\partial^2f}{\partial x^2}-\frac{1}{c^2}\frac{\partial^2f}{\partial t^2}=0
$$
From the geometry alone, it was only needed to note that a change in $t$ multiplied by the velocity yields the same results (as measured by the second derivative) as a change in $x$ --that is, a translation of $f(x-ct)$. See, for example, for a good discussion: kiskis.physics.ucdavis.edu/landau/phy9hc_03/wave.pdf.
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
The "intuition" here is that the wave equation is the equation for a general "disturbance" that has a left- and a right-travelling component, i.e. spreads without any preferred direction given by the equation of motion.
Observe that $$ \left(v^2 \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial t^2}\right) y = 0$$ can be factored as (which is what you probably mean by "squaring" in the question) $$ \left(v\frac{\partial}{\partial x} + \frac{\partial}{\partial t}\right)\left(v\frac{\partial}{\partial x} - \frac{\partial}{\partial t}\right) y = 0$$ implying $$ \left(v\frac{\partial}{\partial x} + \frac{\partial}{\partial t}\right)y = 0 \quad\vee\quad \left(v\frac{\partial}{\partial x} - \frac{\partial}{\partial t}\right) y = 0$$ where it is easy to see that $y(x,t) \equiv y(x - vt)$ and $y(x,t) \equiv y(x + vt)$ are solutions, respectively. Since $v$ is assumed positive, $x-vt$ becomes smaller as the time $t$ passes (whatever $y$ describes is travelling to the right in the usual coordinate system), and similarily, $x+vt$ is travelling to the left. By linearity, the general solution is a sum of left- and right-movers, i.e. $y(x,t) \equiv y_R(x - vt) + y_L(x + vt)$.
The initial conditions $y(x,0) = f(x)$ and $\frac{\partial}{\partial t}y(x,0) = g(x)$ for arbitrary functions of position $f,g$ fully specify the solution by d'Alembert's formula: $$ y(x,t) = \frac{1}{2}\left[f(x-vt) + f(x+vt) + \frac{1}{v}\int_{x-vt}^{x+vt}g(z)\mathrm{d}z\right]$$ where (roughly) $f$ corresponds to the shape of the disturbance and $g$ to the way it will spread. Note that, in particular, for $g = 0$ and $f = \sin$, we obtain just a standing wave.