= Solution
Take the dot here to mean $d/d\tau$, as in the source. Differentiate $u^\alpha=\gamma(c,\mathbf v)$ and the given <spin> components. Their contraction simplifies because the terms involving $\dot\gamma$ cancel:
$$
s_\lambda\dot u^\lambda
=-\gamma(\mathbf s\cdot\mathbf v)\dot\gamma
+\gamma\mathbf s\cdot(\dot\gamma\mathbf v+\gamma\dot{\mathbf v})
=\gamma^2\mathbf s\cdot\dot{\mathbf v}.
$$
The spatial part of the transport equation consequently reads
$$
\frac d{d\tau}(\gamma\mathbf s)
=\frac{\gamma^3}{c^2}(\mathbf s\cdot\dot{\mathbf v})\mathbf v.
$$
Use $\dot\gamma/\gamma=\gamma^2\mathbf v\cdot\dot{\mathbf v}/c^2$ and divide by $\gamma$. The result is
$$
\boxed{\dot{\mathbf s}=\frac{\gamma^2}{c^2}
\left[(\mathbf s\cdot\dot{\mathbf v})\mathbf v
-(\mathbf v\cdot\dot{\mathbf v})\mathbf s\right].}
$$
No assumption of acceleration perpendicular to the velocity was made. In particular the second term cannot be dropped for a particle speeding up along a straight line. This $\mathbf s$ is the three-vector in the source's particular <spin four-vector> parametrization; its Euclidean magnitude need not be the invariant proper-spin magnitude.
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