= Solution
Fix the orientation of the <Berezin integral> by defining its measure as the linear functional that extracts the coefficient of $\prod_{i=1}^N(\theta_i\bar\theta_i)$; all monomials of lower degree integrate to zero. In particular $\int\mathcal D(\bar\theta,\theta)\prod_i(\theta_i\bar\theta_i)=1$. This specifies the sign convention in the notation $d^N\bar\theta\,d^N\theta$.
The <Grassmann change-of-variables formula> gives
$$
d^N\theta'=\det(U)^{-1}d^N\theta,\qquad d^N\bar\theta'=\det(U^*)^{-1}d^N\bar\theta.
$$
For a <unitary matrix>, $\det U\det U^*=1$, so the combined measure is invariant. Diagonalize the <Hermitian matrix> $B$ by a unitary change of variables. If its eigenvalues are $b_i$, the integrand factors as $\prod_i(1-b_i\bar\theta_i\theta_i)=\prod_i(1+b_i\theta_i\bar\theta_i)$. Extracting its top coefficient proves the <Grassmann Gaussian integral>
$$
\boxed{\int d^N\bar\theta\,d^N\theta\,e^{-\bar\theta_iB_{ij}\theta_j}=\prod_i b_i=\det B.}
$$
The determinant identity also holds without Hermiticity by direct expansion; Hermiticity makes the diagonalization proof immediate.
Let $C(x)=\sum_kx_kC^{(k)}$. Integrating out the <Grassmann variables> first gives the exact polynomial expression
$$
I(g)=\int_{\mathbb R^N}d^Nx\,e^{-x^TAx/2}\det[B-gC(x)].
$$
The requested factorization requires \b[$B$ to be invertible], an assumption omitted from the question. Under it, use the formal perturbative <matrix logarithm> near $g=0$:
$$
\det[B-gC(x)]=\det B\,\exp\left\{\operatorname{Tr}\log[I-gB^{-1}C(x)]\right\}.
$$
Consequently
$$
\boxed{V_{\rm eff}(x)=-\operatorname{Tr}\log[I-gB^{-1}C(x)]=\sum_{m\geq1}\frac{g^m}{m}\operatorname{Tr}[B^{-1}C(x)]^m.}
$$
Expanding $C(x)$ gives exactly the displayed coefficients $V^{(m)}_{k_1\ldots k_m}=(m-1)!\operatorname{Tr}[B^{-1}C^{(k_1)}\cdots B^{-1}C^{(k_m)}]$, and hence $I(g)=\det B\int d^Nx\,e^{-x^TAx/2-V_{\rm eff}(x)}$. The logarithmic representation is a formal series, not a globally convergent logarithm for every $x$: the determinant can have zeros. The original determinant integral remains well-defined for singular $B$, but $I(0)=0$ and the requested normalized expansion then cannot be formed.
There is a second convention to specify. The $x_i$ are real, so a complex Hermitian $A$ gives $x^TAx=x^TA_Rx$, where $A_R=\operatorname{Re}A=(A+A^T)/2$ is real symmetric positive definite. Its Gaussian <covariance matrix> is $A_R^{-1}$. It equals $A^{-1}$ only if $A$ is real symmetric. Thus
$$
I(0)=\det B\,(2\pi)^{N/2}(\det A_R)^{-1/2},\qquad\frac{I(g)}{I(0)}=\mathbb E_{x\sim N(0,A_R^{-1})}[e^{-V_{\rm eff}(x)}].
$$
The effective <Feynman rules> are: a bosonic contraction joins indices $k,\ell$ with $(A_R^{-1})_{k\ell}$; an $m$-leg vertex carries $-g^m\operatorname{Sym}V^{(m)}_{k_1\ldots k_m}$; sum all closed <vacuum diagrams> with their <symmetry factors>, including disconnected diagrams. The symmetrization is necessary because the $x_k$ commute: the trace tensor is cyclically invariant but need not be fully symmetric. The minus sign comes from expanding $e^{-V_{\rm eff}}$. Taking the logarithm of the normalized integral instead keeps only connected <vacuum diagrams>.
Equivalently, retain the fermions: an oriented contraction is $(B^{-1})_{ij}=\langle\theta_i\bar\theta_j\rangle$, the mixed vertex has weight $+gC^{(k)}_{ij}$, and every closed <fermion loop> gives a minus sign. The same bosonic covariance joins its $x$ legs. Expanding these contractions recovers the trace vertices above. This is <bosonic trace vertices from a finite fermionic determinant>.
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