= Solution
In the forward direction $\mathbf k_f=\mathbf k_i$, the first Born amplitude $-C\int V$ is real. Symmetrize the second-order double integral under $\mathbf r\leftrightarrow\mathbf r'$ before taking its imaginary part. This replaces $e^{-i\mathbf k_i\cdot\mathbf d}$ by $\cos(\mathbf k_i\cdot\mathbf d)$, and gives
$$
\operatorname{Im}f^{(2)}(0)=C^2\int d^3r\,d^3r'\,
V(\mathbf r)V(\mathbf r')\frac{\sin kd}{d}
\cos(\mathbf k_i\cdot\mathbf d).
$$
The preceding total-cross-section integral equals the same real integral multiplied by $4\pi/k$. Therefore
$$
\boxed{\operatorname{Im}[f^{(1)}(0)+f^{(2)}(0)]
=\frac{k\sigma^{(1)}_{\rm total}}{4\pi}+O(V^3)}.
$$
This is the <optical theorem at second Born order>. Comparing only the real first-order forward amplitude with a nonzero second-order <scattering cross-section> would mix perturbative orders. Terms of order $V^3$ require both the next amplitude order and interference in the <scattering cross-section>.
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