Doesn't the work $W = int F , dx$ count only the work done by the outermost point of a spring?

The restoring force doesn't depend on the relaxed length of the spring or number of particles in it. It just depends on how far it is stretched or compressed with respect to the mean position and the spring constant.

When finding the work done (potential energy stored) we just need the force which we must apply to stop the spring from relaxing, at a certain distance and hence we just need the force at that point and the small displacement $dx$ over which it acts. Finally we integrate it all to get the total work done.

Why don't we sum (integrate) the work done by all the points?

In essence we are doing just that. When we take the integral of $(kx)dx$ between zero (the relaxed position of the end of the spring) to $x$, the stretched position, we are macroscopically summing up the work done on the microscopic level to pull the molecules apart at each point along the length of the spring in opposition to the electrostatic forces between them.

Envision the spring being comprised of molecules connected together by microscopic springs between them. The spring constant, $k$, is the macroscopic parameter equivalent to the microscopic spring constant. The integral of $(kx)dx$ sums up all the work being done on the microscopic level in stretching the microscopic springs.

But then are summing work done on each point only for distance dx

What I mean is the sum of all work at the molecular level equals the total work at the macroscopic level, and the sum of all the intermolecular displacements, $dx_i$, add up to the final displacement, $x$ as @Farcher showed pictorially. Or

$$sum_{i=0}^{n}k_{i}dx_{i}=frac{kx^2}{2}$$

Where $sum dx_{i}$ = $x$.

Hope this helps.

Normally in physics problems the spring is considered to be an ideal spring, with negligible mass and negligible internal energy dissipation. So although the various parts of the spring do move as it is compressed and expanded, if you to work out the $W=Fd$ work associated with all of those various individual parts of the spring, you would realize that the force $F$ associated with moving each of those parts is zero, so no work is done on those parts. You only have to concern yourself with the work done on whatever the ideal spring is pushing.

If you were to ever encounter a physics problem in which a compressed spring is, for example, embedded in some very thick medium so that there is a force associated with the movement of every element of the spring itself as it is released, then one would have to do as you suggested and figure out the $W=Fd$ work associated with every element of the spring and add them all up to get the total work.

Suppose that there is a spring fixed at one end and a force $F$ is applied to the spring.
The important thing is that force $F$ can be thought of as the "tension" in the spring and is the same throughout the spring.

Now consider the whole spring as three equal lengths.

Each length is subjected to a force $F$ at each end as shown in the diagram.

The work done on spring 1 by the forces stretching it by a small amount $delta x$ is $F,delta x$.
The work done on spring 2 by the forces stretching it by $delta x$ is $F,2delta x - F,delta x = F,delta x$.
The work done on spring 3 by the forces stretching it by $delta x$ is $F,3delta x - F,2delta x = F,delta x$.

So the total work done in stretching the three springs is $F,delta x +F,delta x+F,delta x = F,3delta x = FDelta x$ where $Delta x (=3delta x)$ is the total extension of the whole spring.

The whole spring can be cut up into as many segments as you like and then the work done extending the spring can be written as $displaystyle int_0^{x_{rm final}} F,dx$ where $x_{rm final}$ is the final extension of the whole spring.

The other answers are all good, but I love this question too much to pass it up:

Let's break our spring with constant $k$ into $n$ little springs with spring constant $nk$ (shown here in ASCII):

|~$nk$~|~$nk$~|$...$|~$nk$~|~$nk$~|~$nk$~|

You can verify that the total spring constant is indeed $k$:

$$frac 1 {frac 1{ nk} +ldots+frac 1{nk} }= frac 1 {frac n {nk}}=k$$

since series springs are added harmonically. The surprising result is that the spring segments have a larger spring constant than the total spring.

When the total spring is stretched by $x$, each segment stretches by $x/n$.

Note that the tension in each segment is:

$$ T = (nk)(x/n) = kx $$

as it must be. Moreover, each segment has done work:

$$Delta W = frac 1 2 (nk)big(frac x nbig)^2 = frac 1 2 frac {kx^2} n$$

so obviously:

$$W = ntimesDelta W = frac 1 2 kx^2 $$

In the limit $nrightarrow infty$ you can see that the energy is uniformly distributed along the length of the spring.

We're calculating the work done by the spring on some external point which isn't part of the spring. When you calculate the work done on some part of the spring, you're calculating how this contributes to the total kinetic energy of the spring itself.

The short answer to the question in the title is, "Yes, but the work done by that part of the spring on what?". In particular, if the integral is positive, that means energy was transferred 'to' an object, negative means 'from.' What you're asking is the same as asking, "Where did that energy come from before it got to the end of the spring?" As you correctly intuited, that energy gets stored all along the entire spring, but it is in a more advanced class on elastic media you'll learn how to handle that.

In the mean time, if you want to think about it on your own, explore the topic of using multiple springs to build a single "spring." For example, "If I connect two springs with constant $k$ end to end, what is the constant $k_{mathrm{effective}}$ of the combination?" or the same question if I connect two springs side by side.