 Solution

Take $\mathbf{R}={\mathbf{r}}_2-{\mathbf{r}}_1$ and label each dumbbell once. Locally assume its probability density and number density vary negligibly over a connector length. The oriented connector crossing count follows geometrically: a connector crosses a small plane element precisely when its midpoint lies within a slab of normal thickness $|\mathbf{R}\cdot \mathbf{n}|$. For configurations in $d^{3}R$, the unsigned crossing number per unit area is thereforeThe expression without the absolute value is an oriented crossing count, or an unsigned count restricted to $\mathbf{R}\cdot \mathbf{n}\ge 0$. It cannot be an unrestricted nonnegative number: a connector $\mathbf{R}=-R\mathbf{n}$ gives a negative printed expression despite physically intersecting the plane.

To obtain traction, keep the force orientation as well. For $\mathbf{R}\cdot \mathbf{n}>0$, the spring force on the bead below the plane exerted by the bead above it is $H\mathbf{R}$. For the opposite orientation it is $-H\mathbf{R}$. Multiplying that force by the unsigned crossing count in either case givesIntegrating over all labelled configurations yieldsThere is no extra factor of two: each dumbbell was labelled and counted once, and the orientation-dependent force has already supplied the sign. This is the local continuum stress approximation; resolving appreciable spatial variation across a finite connector would require a nonlocal stress construction.