I think some of your confusion stems from the fact that there are two different kinds of vacuua in QFT. First there is the vacuum of the free theory, usually denoted $|0\rangle $, second there is the full (interacting) vacuum, usually denoted $|\Omega \rangle$.
What we want to calculate are the different quantities in the full theory like:
\begin{equation}
\langle\Omega\vert T\phi(x_1)\phi(x_2)\vert\Omega\rangle.
\end{equation}
What you have written in equation (3) is acutally the propagator in the free theory (see that you use $\mathcal{L}_0$ and not $\mathcal{L}$ as it should be). This may seem like a minor point but the difference is essential to understand. (You must have copied it wrong from eqn (9.18) in P & S, since it looks correct there.
The problem is that we do cannot calculate this directly. What we do is we basically say: What if the full theory is almost like the free theory but with a small interaction term? That is, we do perturbation theory. This is because we know the propagator of the free theory:
$$ \Delta_F(x_1-x_2) =\langle0\vert T\phi(x_1)\phi(x_2)\vert0\rangle = \frac{\int \mathcal{D}\phi\; \phi(x_1)\phi(x_2) e^{iS_0}}{\int \mathcal{D}\phi\; e^{iS_0}}.$$
The trick is then to rewrite everything in terms of the propagator of the free theory.
The actual derivation of the full 2-point function is fairly involved, but you can find it in P & S pages 82-99.
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
Perhaps it would be more symmetric if we write the denominator $\langle\Omega|\Omega\rangle=1$ on the left-hand sides explicitly. The denominators on the right-hand sides are important, and cannot in principle be omitted in later analysis.
If we start with eq. (4.31), here the denominator on the right-hand side is crucial, since when we apply the theorem of Gell-Mann and Low a non-trivial factor is cancelled. Here $|\Omega\rangle$ and $|0\rangle$ denote the vacuum in the interaction and the free theory, respectively.
Similarly, in eq. (9.18) the denominator is a convenient way to absorb an over-all normalization factor in the path integral measure.
Eq. (9.27) refers to a free theory, so the Heisenberg and interaction pictures coincide.