So I would have thought that this would be how you derive the work on a spring: basically the same way you do with gravity and other contexts, use $$W=\vec{F}\cdot \vec{x}.$$ If you displace a spring by $x$, then it exerts a force $-k x$, so $F=-kx$, since the displacement is $x$.
So $$W=-kx^2.\qquad \leftarrow\text{ (however, apparently wrong!)}$$
I've seen the correct derivation of work in a spring (with an extra half) and don't doubt that it's correct, but also don't see where my logic fails in this alternate derivation.
Answer
You may be imagining that if you push with constant force $F$, the spring will compress until the spring has such a resistive force.
But since the spring was not counteracting that force, your constant force $F$ was accelerating the mass. Upon reaching the point where the spring has force $F$ as well, the mass does not stop but has a speed such that $KE = \frac{1} {2} k x^2$. So the work your hand does (when pushing with constant force) is correct, but half of the work is put into the spring potential energy and the other half is put in kinetic energy of the mass.
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