I think that the discussion related to some answers and the original question revolves around the proper definition of causality in physical theories (I won't touch here the philosophical issues connected to this concept).
In physics, we say that a time-dependent quantity $A(t)$ has a causal relation with another time-dependent quantity $B(t)$, if we have a theory connecting $A(t)$ to $B(t')$ for all the times $t>t'$. In such a case, we say that $A$ depends causally on $B$.
The temporal sequence between cause and effect has a central role. This implies that a relation at equal times, $A(t)=B(t)$ can never be seen as causal.
Before getting the work-energy theorem, let me discuss some other examples of causal and not-causal relations in classical mechanics as a preliminary exercise.
The Second Law of Newton's dynamics is not a causal relationship between force ($F(t)$) and the acceleration $a(t)$ of a body subject to that force. It is just a formula connecting these two different quantities at the same time. The same holds (notwithstanding some widespread misconception) for all the action-reaction pairs of forces of the Third Newton's law. Again the fact that force $F_{12}(t)$ of the body $2$ on body $1$ is always equal and opposite to the force $F_{21}(t)$ of the body $1$ on body $2$, at the same time, does not imply a causal relation. In Newtonian dynamics, it is a property of the system of two bodies valid at each time $t$ in a completely symmetric way.
The Second Law of Newton's dynamics implies a causal relation between the velocity (or position) at time $t$ and the acceleration (then the force) at a previous time $t'<t$. This is formally evident by writing the solution of the differential equation for the velocity $v(t)$ as
$$
v(t)= v(t') + \int_{t'}^t a(\tau) {\mathrm d} \tau.
$$
I find it even clearer a discretized version of the previous equation:
$$
v(t+\Delta t) = v(t) + a(t) \Delta t + O(\Delta t^2).
$$
Now the discussion of the work-energy theorem in terms of causality should be simple.
The theorem states that the work done from time $t_0$ to time $t$ by the resulting force ${\bf F}$ along the trajectory followed by a particle of mass $m$, moving under the action of that force, is equal, at each time t, to the difference of kinetic energy at time $t$ and time $t_0$:
$$
W_{t_0}(t)=\int_{t_0}^t {\bf F}(\tau)\cdot {\bf v}(\tau) {\mathrm d}\tau=\int_{t_0}^t m \frac{d{\bf v}(\tau)}{dt}\cdot {\bf v}(\tau) {\mathrm d}\tau=\frac12 m \left( v^2(t)- v^2(t_0)\right)=\Delta K_{t_0}(t).
$$
It is clear that, for any choice of the initial time, we have the equality of two functions of the time at the same time. Therefore, based on the previous discussion, we do not have a causal connection between $K_{t_0}(t)$ and $W_{t_0}(t)$.
