The commutator \( [F, T_2] \) consists of two-body and one-body terms. Let us compute first the two-body term, as it results from a single contraction (i.e. a single application of \( [a_p, a^\dagger_q] = \delta_p^q \)). We denote this as \( [F, T_2]_{2b} \) and find
$$ \begin{align*} [F, T_2]_{2b} &= \frac{1}{4}\sum_{pq}\sum_{rsuv} f_p^q t_{ij}^{ab}\left[a^\dagger_q a_p, a^\dagger_a a^\dagger_b a_j a_i \right]_{2b} \\ &= \frac{1}{4}\sum_{pq}\sum_{abij} f_p^q t_{ij}^{ab}\delta_p^a a^\dagger_q a^\dagger_b a_j a_i \\ &- \frac{1}{4}\sum_{pq}\sum_{abij} f_p^q t_{ij}^{ab}\delta_p^b a^\dagger_q a^\dagger_a a_j a_i \\ &- \frac{1}{4}\sum_{pq}\sum_{abij} f_p^q t_{ij}^{ab}\delta_q^j a^\dagger_a a^\dagger_b a_p a_i \\ &+ \frac{1}{4}\sum_{pq}\sum_{abij} f_p^q t_{ij}^{ab}\delta_q^i a^\dagger_a a^\dagger_b a_p a_j \\ &= \frac{1}{4}\sum_{qbij}\left(\sum_{a} f_a^q t_{ij}^{ab}\right)a^\dagger_q a^\dagger_b a_j a_i \\ &- \frac{1}{4}\sum_{qaij}\left(\sum_{b} f_b^q t_{ij}^{ab}\right)a^\dagger_q a^\dagger_a a_j a_i \\ &- \frac{1}{4}\sum_{pabi}\left(\sum_{j} f_p^j t_{ij}^{ab}\right)a^\dagger_a a^\dagger_b a_p a_i \\ &+ \frac{1}{4}\sum_{pabj}\left(\sum_{i} f_p^i t_{ij}^{ab}\right)a^\dagger_a a^\dagger_b a_p a_j \\ &= \frac{1}{2}\sum_{qbij}\left(\sum_{a} f_a^q t_{ij}^{ab}\right)a^\dagger_q a^\dagger_b a_j a_i \\ &- \frac{1}{2}\sum_{pabi}\left(\sum_{j} f_p^j t_{ij}^{ab}\right)a^\dagger_a a^\dagger_b a_p a_i . \end{align*} $$