= Solution
Write the centre position as $\mathbf x$ and the bead positions as $\mathbf r_{1,2}=\mathbf x\mp\mathbf R/2$. In the <affine velocity field>, $\mathbf u(\mathbf r_{1,2})=\mathbf u(\mathbf x)\mp K\mathbf R/2$. The midpoint follows the flow, so its two slip <velocities> are
$$
\dot{\mathbf r}_{1,2}-\mathbf u(\mathbf r_{1,2})=\mp\frac12(\dot{\mathbf R}-K\mathbf R).
$$
Each bead loses mechanical power $\zeta$ times its squared slip <speed>. Adding the two powers proves the <drag dissipation of a bead-spring pair>:
$$
\boxed{D_{\rm pair}=\zeta\left|\frac{\dot{\mathbf R}-K\mathbf R}{2}\right|^2+\zeta\left|\frac{\dot{\mathbf R}-K\mathbf R}{2}\right|^2=\frac\zeta2|\dot{\mathbf R}-K\mathbf R|^2.}
$$
The factor one half comes from two equal-and-opposite bead slips, not from assigning only one bead a drag force.
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