Where did the work done by smaller force go? Suppose I have a spring of spring constant 150 N/m.One person pulls with it a force of 15 N. The extension produced is 0.1 m. Now another person comes and pulls with a force of 30 N (The first person is still there). The final extension is 0.3 m.The initial and final potential energies of the spring are 0.75 J and 6.75 J respectively. The work done by the bigger force is 6 J which is exactly equal to the change in P.E of the spring. So where did the work done by the smaller force (when they were pulling it together) go? Why didn't it help to further increase the potential energy of the spring?
 A: You are trying to compare the work done by constant forces to the change in potential energy of the spring. However, if you take a spring at rest at its current equilibrium and then apply a purely constant force to it, then the spring will not come to rest at a new equilibrium position; it will start oscillating about that new equilibrium instead. This is why, for example, with your first person $\frac12kx^2\neq Fx$ where $x$ is the distance between the old and new equilibrium positions. The constant force does more work getting to the new equilibrium position than the spring does. Therefore, we will still have motion, as the extra energy will have gone into kinetic energy.
Therefore, you need to change either your assumptions or your analysis here. Either make the applied forces variable so that you can stop the spring at any equilibrium as done in Dale's answer, or keep the applied forces constant and then either correctly include kinetic energy, or look at the points where the spring would actually momentarily come to rest (so then you can equate $\frac12kx^2$ and $Fx$).

I will go ahead and provide the other option, choosing the scenario where we apply constant forces, and the spring comes to rest so that we do not need to consider kinetic energy. For a constant applied force $F$, the spring will come to rest when the work done by the spring is equal to the work done by the force, so that the stopping position is $x=2F/k$. Therefore, when the $15\,\mathrm N$ force is applied, the spring will momentarily stop at $x_1=0.2\,\mathrm m$. You can do a similar argument to show that once we then start the $30\,\mathrm N$ force we will get another momentary stopping at $x_2=0.4\,\mathrm m$.
Following Dale's graphs, we have for the forces

and for the work done by each force

As you can see, once the spring momentarily comes to rest at $0.4\,\mathrm m$, the total $12\,\mathrm J$ of work done by the applied forces is equal in magnitude to the work done by the spring. Additionally, each force does $6\,\mathrm J$ of work, so the work done by both people are accounted for in the total work.
A: 
The work done by the bigger force is 6J which is exactly equal to the change in P.E of the spring. 

Actually, this is incorrect. If we don't want to be adding kinetic energy and worrying about that, then the total force from the people must at all times equal the force from the spring. This means that the force from the second person is only 30 N at full extension and is less than 30 N before then. Here is a plot of the forces vs displacement. The blue line is the total force from both people, the yellow line is the force from the first person, and the green line is the force from the second person.

In this plot the work is the area under the curve. It is immediately obvious that the area under the blue curve is the sum of the areas under the curve for the yellow and for the green person.
What is not so obvious is that the work done by the first person is actually more than the work done by the second. The area under the green curve is 3.0 J, not 6.0 J. The area under the yellow curve is 3.75 J, with 0.75 J under the initial triangular portion from 0 to 0.1 m and the remaining 3.0 J in the flat section.
So, indeed, the work done by the smaller force is essential for calculating the total energy that went into the spring. The calculation of 6.0 J from the second person is simply incorrect. It is incorrect by a factor of 2 which basically is the difference between incorrectly assuming that the 30 N force was applied throughout the movement instead of correctly recognizing that it is only 30 N at full extension. Of course, alternatively you could consider the change in KE that would result from a constant 30 N force, as described by @Aaron Stevens in his answer.
A: When both are pulling it together, the total fore is $45\ \mathrm N$ and the extension is $0.3\ \mathrm m$ and the work-done is still $6.75\ \mathrm J$. When they pull individually the work done is $6\ \mathrm J$ and $0.75\ \mathrm J$ respectively irrespective of who pulls first. It doesn't matter how they are being pulled; together or individually, the contribution remains the same.
