= Solution
For one <Grassmann variable>, define the <Berezin integral> by $\int d\theta\,1=0$ and $\int d\theta\,\theta=1$, extending linearly. Since $\theta^2=0$, these rules determine it on every polynomial. With several <Grassmann variables>, integrate from the rightmost differential first. Equivalently, $\int d\theta_n\cdots d\theta_1$ extracts the coefficient of $\theta_1\cdots\theta_n$ in that order. Repeated indices give zero, while permuting distinct generators gives the <fermionic sign> of that permutation. Hence
$$
\int d\theta_n\cdots d\theta_1\,\theta_{i_1}\cdots\theta_{i_n}=\varepsilon_{i_1\cdots i_n},\qquad\varepsilon_{1\cdots n}=1,
$$
where $\varepsilon$ is the <Levi-Civita symbol>.
For the doubled variables, fix the paired measure orientation by
$$
\int\prod_{i=1}^n d\bar\theta_i\,d\theta_i\;\theta_1\bar\theta_1\cdots\theta_n\bar\theta_n=1.
$$
The exponential in the <Grassmann Gaussian integral> has a finite Taylor expansion. Only its term of degree $2n$ survives integration. Each barred and unbarred index must occur exactly once; the $n!$ orders of the even bilinears cancel the Taylor factorial. Reordering the unbarred indices gives the determinant permutation sign. Therefore
$$
Z(B)=\int\prod_i d\bar\theta_i\,d\theta_i\,e^{-\bar\theta B\theta}=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_iB_{i\sigma(i)}=\boxed{\det B}.
$$
This identity holds even for a singular <matrix>. The normalized expectations below require $B$ invertible.
Put $C=B^{-1}$. Differentiating with respect to $B_{ji}$ inserts $-\bar\theta_j\theta_i=\theta_i\bar\theta_j$. The <derivative of a determinant> then gives the <ordered Grassmann Gaussian contractions>
$$
\boxed{\langle\theta_i\bar\theta_j\rangle=\frac1{\det B}\frac{\partial\det B}{\partial B_{ji}}=C_{ij}.}
$$
Differentiate once more, using $\partial C_{ij}/\partial B_{lk}=-C_{il}C_{kj}$. The result is
$$
\boxed{\langle\theta_i\bar\theta_j\theta_k\bar\theta_l\rangle=C_{ij}C_{kl}-C_{il}C_{kj}.}
$$
The minus sign is the interchange sign in the fermionic <Wick theorem>. In particular $\langle\bar\theta M\theta\rangle=-\operatorname{tr}(MC)$. Reordering each barred-unbarred pair in the product and contracting the indices gives
$$
\boxed{\langle(\bar\theta M\theta)(\bar\theta N\theta)\rangle=\operatorname{tr}(MC)\operatorname{tr}(NC)-\operatorname{tr}(MCNC).}
$$
The <covariance of Grassmann bilinears> is the connected second term. Its <Feynman diagram> is one oriented <fermion loop> through both insertions; the first term has two separate <fermion loops>, one at each insertion. Each closed <fermion loop> contributes a minus sign, so two loops give a plus and one gives a minus.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-51-fermion-loops.png]
{title=Connected and disconnected contractions of Grassmann bilinears}
For the <Dirac field>, write $D=\gamma\cdot\partial+m$ and use signature $(-,+,\ldots,+)$, consistent with the denominator in the question. A finite regulator turns the free <fermionic path integral> into the same Gaussian problem with $B=iD$, because $e^{iS}=e^{-i\bar\psi D\psi}$. The <time-ordered product> covariance is consequently $(iD_F)^{-1}=-iD_F^{-1}$, where the subscript specifies the <Feynman i-epsilon prescription>.
With the <Fourier transform> convention $\psi(x)=\int d^dp\,(2\pi)^{-d}e^{ip\cdot x}\widetilde\psi(p)$, the derivative kernel is $i\gamma\cdot p+m$. The <gamma matrices> obey
$$
(i\gamma\cdot p+m)(-i\gamma\cdot p+m)=(p^2+m^2)I,
$$
since their anticommutator makes $(\gamma\cdot p)^2=p^2I$. Thus the <mostly-plus Dirac propagator with a real derivative kernel> is
$$
\boxed{\langle T\psi(x)\bar\psi(0)\rangle=iS_F(x),\qquad\widetilde S_F(p)=\frac{-i\gamma\cdot p+m}{-p^2-m^2+i0}.}
$$
For $p^2=-p_0^2+\mathbf p^2$, the denominator is $p_0^2-\mathbf p^2-m^2+i0$. Its positive-energy pole is below and negative-energy pole above the real axis, giving <time ordering>. Here $S_F$ is defined without the leading factor $i$, as in the question; conventions that include that factor in the name <Dirac propagator> give the same covariance.
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