= Solution
Use incoming <fermion> $p_1$ and <antiparticle> $p_2$, outgoing <fermion> $p_3$ and <antiparticle> $p_4$. The two <tree-level Feynman diagrams> are the $t$ channel, with $q_t=p_1-p_3=p_4-p_2$, and the $s$ channel, with $q_s=p_1+p_2=p_3+p_4$. They occupy the lower row of the figure. In the $t$ channel the <fermion flow> follows $1\to3$ and $4\to2$; in the $s$ channel it follows $1\to2$ at the incoming <interaction vertex> and $4\to3$ at the outgoing <interaction vertex>.
The two contractions again have a relative <fermionic sign>. With the same harmless common-phase convention as <2/b/i>, their sum is
$$
\boxed{\mathcal M_{f\bar f}=g^2\left[(\bar u_3u_1)(\bar v_2v_4)-(\bar v_2u_1)(\bar u_3v_4)\right].}
$$
The <auxiliary-field propagator> is the same constant $-i$ in both channels. There is no $u$ channel preserving the external <fermion number> assignments.
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