= Solution
Use a Minkowski <external source> term $+\int J\chi$, where $\chi$ is the integration variable, and normalize the vacuum functional:
$$
Z[J]=\frac{\int\mathcal D\chi\,\exp\{iS[\chi]+i\int d^4x\,J(x)\chi(x)\}}
{\int\mathcal D\chi\,e^{iS[\chi]}},\qquad W[J]=\log Z[J].
$$
The usual vacuum boundary prescription is implicit. This is the logarithmic convention $Z=e^W$ for the <connected generating functional>. It makes the given quadratic expression consistent with $\Delta_F=\langle T\chi\chi\rangle$. Another common convention is $Z=e^{iW_c}$, with $W_c=-iW$; mixing the two conventions would change the factors of $i$ below.
<Functional derivative> generates the full and connected <correlation functions>:
$$
G_n(x_1,\ldots,x_n)=\left.i^{-n}\frac{\delta^n Z}{\delta J(x_1)\cdots\delta J(x_n)}\right|_{J=0},
\qquad
G_n^c(x_1,\ldots,x_n)=\left.i^{-n}\frac{\delta^n W}{\delta J(x_1)\cdots\delta J(x_n)}\right|_{J=0}.
$$
These are <time-ordered products> in the vacuum, and $G_1=G_1^c=0$ by the stated assumption. Differentiating $e^W$ four times partitions the four insertions into connected blocks. All partitions with a singleton vanish, leaving
$$
\boxed{G_4(1,2,3,4)=G_4^c(1,2,3,4)
+G_2^c(1,2)G_2^c(3,4)
+G_2^c(1,3)G_2^c(2,4)
+G_2^c(1,4)G_2^c(2,3).}
$$
The shorthand arguments stand for <spacetime> points; this is a functional cumulant identity, not an assumption of a free field.
To construct the <quantum effective action>, define the source-dependent classical field
$$
\varphi(x)=\langle\chi(x)\rangle_J=-i\frac{\delta W}{\delta J(x)}.
$$
Locally invert this relation to obtain $J[\varphi]$, assuming the response <integral kernel> is invertible with the chosen regulator and boundary prescription. Take its <Legendre transform>
$$
\boxed{i\Gamma[\varphi]=W[J[\varphi]]-i\int d^4x\,J[\varphi](x)\varphi(x).}
$$
The chain-rule terms proportional to $\delta J$ cancel, so $\delta\Gamma/\delta\varphi=-J$. A field-independent constant is fixed separately and is immaterial to the vertices. At zero <external source> the classical field is zero, and there is no linear term.
For the two-point vertex, differentiate the classical field once. If $G=G_2^c$, then
$$
\frac{\delta\varphi(x)}{\delta J(y)}=iG(x,y),\qquad
\frac{\delta J(x)}{\delta\varphi(y)}=-iG^{-1}(x,y),
\qquad
\int d^4z\,G(x,z)G^{-1}(z,y)=\delta^{(4)}(x-y).
$$
The vertex coefficients in this problem are <derivatives> of $i\Gamma$, not of $\Gamma$ alone. Therefore
$$
\boxed{\Gamma_2(x,y)=-G^{-1}(x,y).}
$$
For the three-point vertex, the <derivative> of the connected two-point <integral kernel> at a general <external source> is
$$
\frac{\delta G(a,b)}{\delta J(c)}=iG_3^c(a,b,c).
$$
Combine this with $\delta J/\delta\varphi=-iG^{-1}$ and differentiate the inverse identity, $\delta G^{-1}=-G^{-1}(\delta G)G^{-1}$. This gives the <logarithmic-source inverse-kernel vertices> relation
$$
\boxed{\Gamma_3(x,y,z)=\int d^4a\,d^4b\,d^4c\,
G^{-1}(x,a)G^{-1}(y,b)G^{-1}(z,c)G_3^c(a,b,c).}
$$
All kernels here are finally evaluated at zero <external source>. The connected three-point function is amputated by its three inverse <quantum field theory propagators>.
For the specified quadratic $W$, the <external source> response is $\varphi=i\Delta_FJ$, hence $J=-i\Delta_F^{-1}\varphi$. Substitution in the transform gives
$$
\boxed{\Gamma[\varphi]=\frac i2\int d^4x\,d^4y\,
\varphi(x)\Delta_F^{-1}(x-y)\varphi(y),\qquad
\Gamma_2=-\Delta_F^{-1},\qquad\Gamma_n=0\ (n\geq3).}
$$
For the standard free <quantum field theory propagator> $\widetilde\Delta_F(p)=i/(p^2-m^2+i0)$, this is the usual quadratic free action, up to the vacuum prescription. The <generating functional> is Gaussian, so connected functions beyond second order vanish and there are no interaction vertices or nontrivial <scattering amplitudes>. Full higher even-point functions can still be nonzero: they are products of two-point functions, as the four-point formula shows.
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