= Solution
Take $\mathbf R=\mathbf r_2-\mathbf r_1$ and label each <Hookean dumbbell model>[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 $d^3R$, the \b[unsigned] crossing number per unit area is therefore
$$
\boxed{n|\mathbf R\cdot\mathbf n|\psi(\mathbf x,\mathbf R,t)\,d^3R.}
$$
The expression without the <absolute value> is an oriented crossing count, or an unsigned count restricted to $\mathbf R\cdot\mathbf n\ge0$. 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 <Hooke's law>[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 gives
$$
d\mathbf t=nH\mathbf R(\mathbf R\cdot\mathbf n)\psi\,d^3R.
$$
Integrating over all labelled configurations yields
$$
\mathbf t=\left[nH\int\mathbf R\mathbf R^T\psi\,d^3R\right]\mathbf n,
\qquad
\boxed{\boldsymbol\sigma_{\rm spring}=nH\langle\mathbf R\mathbf R^T\rangle.}
$$
There is no extra factor of two: each <Hookean dumbbell model>[dumbbell] was labelled and counted once, and the orientation-dependent <force> has already supplied the sign. This is the local <Continuum mechanics>[continuum] <Cauchy stress tensor>[stress approximation]; resolving appreciable spatial <Calculus of variations>[variation] across a finite connector would require a nonlocal <stress> construction.
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