Eq. (9.26) is the numerator of eq. (9.18). Using a Fourier decomposition of $\phi(x_i)=\frac{1}{V}\sum_ne^{-ik_n\cdot x_i}\phi(k_n)$, with $k_n^\mu=2\pi n^\mu/L$, we get:
$$\frac{1}{V^2}\sum_{m\, l} e^{-i(k_m\cdot x_1+k_l\cdot x_2)}\left(\prod_{k_n^0>0}\int dRe\phi_n\, dIm\phi_n\right)\left(Re\phi_m+iIm\phi_m\right)\left(Re\phi_l+iIm\phi_l\right)\times\,\times{\rm Exp}\left[-i\frac{1}{V}\sum_{k_n^0>0}(m^2-k_n^2)\left((Re\phi_n)^2+(Im\phi_n)^2\right)\right].\tag{9.26}$$
Note that the parentheses around $\left(\prod \int...\right)$ are essential!
The $\phi$'s that appear above are in momentum space and we use the shortcut $\phi(k_m)=\phi_m$.
I understand now that only the $m=-l$ term in the sum survives. And I also understand that the cross terms ($Im\phi_n Re\phi_l$ and $Im\phi_l\, Re\phi_n$) vanish. Then the expression reduces to:
$$\frac{1}{V^2}\sum_{m} e^{-ik_m(x_1- x_2)}\left(\prod_{k_n^0>0}\int dRe\phi_n\, dIm\phi_n {\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)\left((Re\phi_n)^2+(Im\phi_n)^2\right)\right]\right)\times\\\times\left((Re\phi_m)^2+(Im\phi_m)^2\right)~~~(*)$$
where I have used $k_{-l}=-k_l$ and $\phi(-k_m)=\phi^*(k_m)$ and hence $Re\phi_{-l}=\phi_l$ and $Im\phi_{-l}=-Im\phi_l$.
Next I evaluate the integrals that appea in the product in the equation above. I distinguish the case $n=m$ and $n\neq m$.
For $n=m$ I get:
$$\int dRe\phi_m\, dIm\phi_m\,(Re\phi_m)^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)\left((Re\phi_m)^2+(Im\phi_m)^2\right)\right]\\
+
\int dRe\phi_m\, dIm\phi_m\,(Im\phi_m)^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)\left((Re\phi_m)^2+(Im\phi_m)^2\right)\right]\\
=\int dRe\phi_m\,(Re\phi_m)^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)(Re\phi_m)^2\right]\int dIm\phi_m\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)(Im\phi_m)^2\right]\\
+
\int dIm\phi_m\,(Im\phi_m)^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)(Im\phi_m)^2\right]\int dRe\phi_m\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)(Re\phi_m)^2\right]\\
= 2 \int dx\,x^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)x^2\right]\int dy\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)y^2\right]
$$
For $m\neq n$ I get:
$$\int dRe\phi_n\, dIm\phi_n{\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)\left((Re\phi_n)^2+(Im\phi_n)^2\right)\right]\\
+
\int dRe\phi_n\, dIm\phi_n{\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)\left((Re\phi_n)^2+(Im\phi_n)^2\right)\right]\\
= \left( \int dx {\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)x^2\right]\right)^2
$$
Note that no $Re\phi_m$ and $Im\phi_m$ factors appear, since the $m=n$ case includes these factors already.
Putting everything together eq (*) becomes:
$$(*)=\frac{1}{V^2}\sum_{m} e^{-ik_m(x_1- x_2)}\left(2 \int dx\,x^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)x^2\right]\int dy\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)y^2\right]\right)\times\\
\times\left(\prod_{k_n^0>0, n\neq m}\left( \int dx {\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)x^2\right]\right)^2\right)~~(**)$$
Now it is easy to see that the denominator in (9.18) is:
$$\int D\phi e^{i S_0}=\prod_{k_n^0>0}\int Re\phi_n\, dIm\phi_n\, {\rm Exp}\left[-\frac{i}{V}(m^2-k_n^2\left((Re\phi_n)^2+(Im\phi_n)^2\right))\right]\\
=\prod_{k_n^0>0} \left(\int dx {\rm Exp}\left[-\frac{i}{V}(m^2-k_n^2)x^2\right]\right)$$
This is almost what appears in $(**)$. We manipulate $(**)$ further:
$$(**)=\frac{1}{V^2}\sum_{m} e^{-ik_m(x_1- x_2)}2 \frac{\int dx\,x^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)x^2\right]}{\int dy\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)y^2\right]}\left(\prod_{k_n^0>0}\left( \int dx {\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)x^2\right]\right)^2\right)\\
=\frac{1}{V^2}\sum_{m} e^{-ik_m(x_1- x_2)}2 \frac{\int dx\,x^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)x^2\right]}{\int dy\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)y^2\right]}\,e^{i S_0}~~~(***)$$
I evaluate the remaining integrals by substituting $k_m^2-m^2\to k_m^2-m^2+i\epsilon$. I get for the integral
$$\frac{\int dx\,x^2{\rm Exp}\left[i\frac{1}{V}(k_m^2-m^2+i\epsilon)x^2\right]}{\int dy\,{\rm Exp}\left[i\frac{1}{V}(k_m^2-m^2+i\epsilon)y^2\right]}=\frac{1}{2}\frac{-iV}{ -i\epsilon- k_m^2+ m^2}$$
Inserting this into (***) I arrive at:
$$(***)=\frac{1}{V}\sum_{m} e^{-ik_m(x_1- x_2)}\frac{-i}{ m^2-k_m^2-i\epsilon}e^{i S_0}~~~(****)$$
Now inserting everything into (9.18):
$$(9.18)=\frac{****}{e^{iS_0}}=\frac{1}{V}\sum_{m} e^{-ik_m(x_1- x_2)}\frac{-i}{ m^2-k_m^2-i\epsilon}$$
Finally I use $\frac{1}{V}\sum_{k_n}\to \int \frac{d^4k}{(2\pi)^4}$:
$$<\Omega| T \phi_H(x_1)\phi_H(x_2)|\Omega>= \int \frac{d^4k}{(2\pi)^4}e^{-ik(x_1- x_2)}\frac{i}{ k^2-m^2+i\epsilon}$$
and this is eq (9.27) in Peskin Schoeder.
Best Answer
Integrate by parts twice (for $x$). For simplicity, I omit $m$ and $\epsilon$: \begin{align} & \frac{i}{2} \int d^4y \, \int d^4x \, D_F(x-y) J(y) (-\partial_x^2) \phi'(x) \\ &= - \frac{i}{2} \int d^4y \, \int d^4x \, D_F(x-y) J(y) \partial_x (\partial_x \phi'(x)) \\ &= \frac{i}{2} \int d^4y \, \int d^4x \,(\partial_x D_F(x-y)) J(y) (\partial_x \phi'(x)) \\ &= - \frac{i}{2} \int d^4y \, \int d^4x \, (\partial_x^2 D_F(x-y)) J(y) \phi'(x) \\ &= \frac{i}{2} \int d^4y \, \int d^4x \, (-\partial_x^2) D_F(x-y)) J(y) \phi'(x) \\ &= \frac{i}{2} \int d^4y \, \int d^4x \, i \mathop{\delta^{(4)}}(x-y) J(y) \phi'(x) \\ &= - \frac{1}{2} \int d^4x \, J(x) \phi'(x) . \end{align}