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First fix the propagator kernels in a positive-sign Gaussian generating functional. From its scalar exponent, $i^{-2}{\delta }^2{Z}_0/\delta J\delta J$ gives the contraction $D(x-y)=-i\Delta (x-y)$. The fermionic insertion rules similarly give $F(x-y)=⟨T\psi (x)\bar{\psi }(y){⟩}_0=-iS(x-y)$, writing $S$ for the kernel labelled $S_F$ in that exponent. Thus the kernels in the question do not already include the factors of $i$ in the actual time-ordered contractions.

At order $g^2$, expand the interaction operator to its second term, or equivalently apply Wick's theorem toFor a connected contribution, the external scalar insertions attach to different interaction points. The two assignments are equal after exchanging $x,y$ and cancel $1/2$. The connected fermionic contraction isThe minus sign is the closed fermion loop sign. The alternative pairing gives two separate fermion tadpoles and belongs to disconnected external components, not the requested connected function. A free external scalar line multiplying a vacuum graph also cancels in the normalized result.

Combining $(ig{)}^2=-g^2$ with the loop minus sign givesThere are four contractions, so replacing $D=-i\Delta$ and $F=-iS$ introduces $(-i{)}^4=1$. ThereforeThis is the fermion-loop diagram in the last panel of i. The trace contracts spinor indices around the loop; it is not an additional degeneracy factor. A convention using physical contractions directly obtains the preceding formula in $D,F$ with the same coefficient. Mixing one convention's Gaussian kernels with another's insertion factors would change its apparent signs.