= Solution
First fix the <propagator kernels in a positive-sign Gaussian generating functional>. From its scalar exponent, $i^{-2}\delta^2Z_0/\delta J\delta J$ gives the contraction $D(x-y)=-i\Delta(x-y)$. The fermionic insertion rules similarly give $F(x-y)=\langle T\psi(x)\bar\psi(y)\rangle_0=-iS(x-y)$, 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> to
$$
\frac{(ig)^2}{2}\int d^4x\,d^4y\,
\left\langle T\phi(x_a)\phi(y_a)
\phi(x)\bar\psi(x)\psi(x)\phi(y)\bar\psi(y)\psi(y)\right\rangle_0.
$$
For 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 is
$$
\left\langle T\bar\psi_\alpha(x)\psi_\alpha(x)
\bar\psi_\beta(y)\psi_\beta(y)\right\rangle_{0,c}
=-F_{\alpha\beta}(x-y)F_{\beta\alpha}(y-x)
=-\operatorname{tr}[F(x-y)F(y-x)].
$$
The 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 gives
$$
G^{(2)}_c\big|_{g^2}=g^2\int d^4x\,d^4y\,
D(x_a-x)D(y_a-y)\operatorname{tr}[F(x-y)F(y-x)].
$$
There are four contractions, so replacing $D=-i\Delta$ and $F=-iS$ introduces $(-i)^4=1$. Therefore
$$
\boxed{G^{(2)}_c\big|_{g^2}=g^2\int d^4x\,d^4y\,
\Delta(x_a-x)\Delta(y_a-y)\operatorname{tr}[S(x-y)S(y-x)].}
$$
This is the fermion-loop diagram in the last panel of <1/b/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.
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