What makes a transverse wave maintain its shape? Consider a simple transverse wave propagating along a rope. We understand it's propagation stating that each point is pulled by its neighbors, making it act along with them. This will create a similar motion in the neighboring points. But, why do they follow with exactly the same shape with no distortion (in the absence of damping).
Further, considering again a point only, shouldn't it continue waving even when the wave has passed, as once initiated by the neighboring points it will either have forces acting or having a velocity in the transverse direction which never makes it static?
 A: While answers posted so far have correct mathematical descriptions, I will look at what you are requesting for thinking about what a point on the rope experiences and how it can come to rest after the wave passes.
We will look at a Gaussian pulse traveling down a rope, as shown below

Now, according to the wave equation (which can be derived from thinking about the forces acting on each segment of the rope)
$$\frac{\partial^2 y}{\partial t^2}\propto\frac{\partial^2 y}{\partial x^2}$$
what this means qualitatively is that the acceleration $\partial^2 y/\partial t^2$ of a point on the rope is proportional to the curvature $\partial^2 y/\partial x^2$ of the rope at that point.
Another way to think of this is that the acceleration of a point on the rope is proportional to how it's height $y$ differs from the average of the heights of the pieces around it.$^*$ For example, at a point that is currently the wave peak, its neighboring points are below it, so it's acceleration is downwards. Contrast this to a point that is at the very beginning or very end of the wave. It has one neighbor at its same height (essentially not moving) and one neighbor above it. Therefore, this point will feel an upward acceleration. This causes a piece that is at rest to start moving upwards (beginning of the wave) and a piece that is slowing down to be at rest to continue slowing down (end of the wave).
To illustrate this, let's zoom in on part of this rope as the wave moves by, and let's show the acceleration of each point of the rope by an arrow:

As you can see, the acceleration of a point on the rope depends on the curvature at that point. This explains why a point on the rope stops moving after the wave passes. The point is already moving downwards, and as the wave is finishing passing by the point's acceleration is upwards. Hence it slows down even more. This continues until the point on the rope essentially comes to rest. Since the other points around it have the same $y$ position, there is also no acceleration at future time points, so this point will not move again.

$^*$This idea is explained much better with much better visuals on the 3Blue1Brown video about partial differential equations. In that video, Grant talks about the heat equation
$$\frac{\partial T}{\partial t}=\alpha\frac{\partial^2T}{\partial x^2}$$
so that just the time rate of change of $T$ is proportional to the curvature.
A: The wave equation for a rope along the $x$-axis is:
$$  \frac{\partial^2 y}{dt^2}= c^2\frac{\partial^2 y}{dx^2} $$
A general solution moving in 1 direction is any $f(x, t)$ that can be written:
$$ f(x, t) = f(x - ct)$$
which, by definition, keeps it shape while propagating to larger $x$, at speed $c$, as $t$ increases.
In terms of Fourier analysis, the function $f(x)$ can be projected into various wavenumbers, $k$, that oscillate at frequency:
$$ \omega = ck $$
which gives the phase velocity:
$$ v_{\phi} \equiv \frac{\omega} k = c $$
while the "shape" propagates at the group velocity:
$$ v_g \equiv \frac{d\omega}{dk} = c $$
This is dispersionless propagation, which keeps its shape.
So the question is, what cause wave to not keep their shape? One is damping, and another is dispersion. For example, if the finite thickness and stiffness of the rope are considered, the equation becomes:
$$\frac{\partial^2 y}{dt^2}=c^2\left(\frac{\partial^2 y}{dx^2}-\epsilon \frac{\partial^4 y}{dx^4}\right)$$
which is dispersive. Wave speed increases (slowly) with frequency.
So: from the point of view of a single section of rope, is there an intuitive connection between local forces, that the point can know about, and the preservation of the global shape...about which the point on the rope cannot know?
One would think not, but perhaps there is a connection. The preservation of the solution's shape as a traveling wave comes from the symmetry in the differential equation: both time and space enter at the 2nd order, leading to any function of $x-ct$ solving the equation in a shape preserving traveling wave.
For a locally isolated point on the rope, all that it knows is that its vertical acceleration ($d^2y/dt^2$) is not proportional to its displacement ($y$), nor the slope ($dy/dx$), but the second order term: $d^2y/dx^2$.
A: The real explanation for your question comes from some detailed mathematics having to do with Fourier series and solutions to wave equations, but you seem to want some intuition for what is going on, which is a totally reasonable thing to ask. 
A wave of any shape can be built up from combinations of sine waves of different wavelengths (which is what the Fourier series is). So even a square wave can be built up from sine waves {demo}. The key thing here is that in a rope, the speed of the wave does not depend on the wavelength, which means that as the wave moves along the string, all the Fourier components move at the same speed and stay in phase. 
In a way, you could imagine that after the wave passes and the point on the rope is staying still, all the fourier components of the wave are still there, but they are cancelling each other out. 
