Intuitive

Thomas precession normally occurs when a spinning particle such as an electron orbits around an atomic nucleus (figure 2); this contributes to its effective spin-orbit coupling, which in turn affects spectral lines (see e.g. [41, section 11.8]).https://arxiv.org/abs/1711.05753

Concrete

Abstract

[T]he revolution of the particle around the nucleus can be interpreted as a sequence of Lorentz boosts.[…] Abstractly, the effect is due to Wigner rotations [3] that appear in representations of the Poincaré group, so it is purely kinematical—it follows solely from space-time symmetries. https://arxiv.org/abs/1711.05753

To get the Lorentz transformation mapping from frame 1 to 3 directly, we compose boosts:

$$\exp\left(\frac{a_y}{c} \,\delta t\, K_y + \frac{a_x}{c} \,\delta t\, K_x\right) \exp\left(\frac{v_x}{c}\,K_x\right) = \exp\left(\frac{v_x}{c}\,K_x + \frac{a_x}{c}\,\delta t\,K_x + \frac{a_y}{c}\,\delta t\, K_y + \frac{a_y\,v_x}{2\,c^2}\,[K_y,\,K_x]\,\delta t + \cdots\right)\tag{1}$$

by the Dynkin formula version of the Baker-Campbell-Hausdorff theorem. From the commutation relationships for the Lorentz group we get $[K_y,\,K_x] = i\,J_z$, so that $\frac{a_y\,v_x}{2\,c^2}\,[K_y,\,K_x]\,\delta t$ corresponds to a rotation about the $z$ axis of $a_y\,v_x\,\delta t / (2\,c^2)$ radians (here, somewhat obviously $J_z,\,J_y,\,J_z$ are "generators of rotations", *i.e.* tangents to rotations about the $x,\,y,\,z$ axes at the identity, respectively, whilst $J_z,\,J_y,\,J_z$ are "generators" of boosts).https://physics.stackexchange.com/a/99459/37286

Why is it interesting?

[T]he phenomenon of Thomas precession is one of the best established kinematical consequences of Poincaré symmetryhttps://cqgplus.com/2018/02/15/can-you-see-asymptotic-symmetries/