How does the work-energy theorem relate to the first law of thermodynamics?

You mention spring compression.

The procedure to calibrate a spring is as follows: you compress the spring, and over the entire travel of the spring you measure the force that it exerts. That generates a force-displacement profile. To find the potential energy stored in a spring for any value of the displacement you integrate the force over distance, using the profile you measured. That procedure is the way to arrive at a mathematical expression for the potential energy stored in a spring.

The work-energy theorem gives a relation between force acting over distance and change of kinetic energy.

$$ int_{s_0}^s F ds = tfrac{1}{2}mv^2 - tfrac{1}{2}mv_0^2 qquad (1) $$

The work-energy theorem is - as the name says - a theorem, not a definition.

For completeness the derivation of the work-energy theorem

In the course of the derivation the following two relations will be used:

$$ ds = v dt qquad (11) $$

$$ dv = frac{dv}{dt}dt qquad (12) $$

The integral for acceleration from a starting point $s_0$ to a final point $s$

$$ int_{s_0}^s a ds qquad (13) $$

Use (11) to change the differential from ds to dt. Since the differential is changed the limits change accordingly.

$$ int_{t_0}^t a v dt qquad (14) $$

Rearrange the order, and write the acceleration $a$ as $tfrac{dv}{dt}$

$$ int_{t_0}^t v frac{dv}{dt} dt qquad (15) $$

Use (12) for a second change of differential, again the limits change accordingly.

$$ int_{v_0}^v v dv qquad (16) $$

Putting everything together:

$$ int_{s_0}^s a ds = tfrac{1}{2}v^2 - tfrac{1}{2}v_0^2 qquad (17) $$

Combining with $F=ma$ gives the work-energy theorem

$$ int_{s_0}^s F ds = tfrac{1}{2}mv^2 - tfrac{1}{2}mv_0^2 qquad (18) $$

The work energy theorem (in classical mechanics) is more general than you describe, it applies to any system of particles, even deformable bodies that interact via instantaneous forces, provided work of all forces is accounted for, including work of internal forces. We can formulate the theorem thus:

Sum total of work of all forces, external and internal, on a many particle system, equals change in its kinetic energy.

If we do not account for work of internal forces, like when the body is deformable but we ignore work of force due to one part of the body on the other part, then the work-energy theorem does not hold.

1st law of thermodynamics is similar to work-energy theorem, but it is not "more general". This is because 1) it is about thermodynamic systems, not mechanical systems; 2) it refers to concept of internal energy, which does not include mechanical kinetic energy but does include mechanical potential energy; 3) it says something very different about the internal energy, than the work-energy theorem says about kinetic energy. We can formulate 1st law thus:

For any body there exists a internal energy function $U$ of equilibrium state $X$ which can change via heat transfer $Delta Q$ or work transfer $Delta W$ with the surrounding bodies; since the effect of heat transfer can be equally achieved by some equivalent amount of work transfer, we use the same units for both work and heat: $$ Delta U = Delta W + Delta Q. $$ Notice how the law introduces a new quantity, internal energy. This is very different from the work-energy theorem, which only says how much kinetic energy increases as a result of work.