To be specific, suppose we have $(n,1)$ random vector $x \sim N(\mu, \Sigma)$ where $\mu$ is $(n,1)$ and $\Sigma$ is $(n,n)$. Define: \begin{align*} Y & = x'Ax + b'x + c \end{align*}
Then what is the following (for $t \in \mathbb{R}$)? \begin{align*} E(e^{tY}) \end{align*}
I will build on my answer from here: https://math.stackexchange.com/questions/442472/sum-of-squares-of-dependent-gaussian-random-variables/442916#442916 and use notation from there. First I will look at the case without the linear and constant term, then we will see how to take them into account.
So let $Q(X)=X^T A X$ be a quadratic form in the multivariate normal vector $X$, with expectation $\mu$ and covariance matrix $\Sigma$. We found that $$ Q(X)=\sum_{j=1}^n \lambda_j (U_j+b_j)^2 $$ where $Z=Y-\Sigma^{-1/2}\mu$, we use the spectral theorem to write $\Sigma^{1/2}A \Sigma^{1/2} = P^T \Lambda P$, $P$ orthogonal and $\Lambda$ diagonal with positive diagonal elements $\lambda_j$, and $U=PZ$ so that $U$ has independent standard normal components $U_j$. The we can define $b=P \Sigma^{-1/2} \mu$. To summarize so far, $Q(X)$ is written above as the sum of independent scaled noncentral chisquare random variables. Using https://en.wikipedia.org/wiki/Noncentral_chi-squared_distribution we can see that $(U_j+b_j)^2$ is noncentral chisquare with one degree of freedom and noncentrality parameter $b_j^2$. Then its moment generating function (mgf) is given by $$ M_j(t) = \frac{\exp\left(\frac{t b_j^2}{1-2t} \right)}{(1-2t)^{1/2}} $$ Then we find the mgf of $\lambda_j (U_j+b_j)^2$ as $M_j(\lambda_j t)$, and the mgf $M(t)$ of the sum $Q(X)$ the product of this: $$ M(t) = \frac{\exp\left(\sum_{j=1}^n \frac{b_j^2 \lambda_j t}{1-2t\lambda_j} \right)}{\exp(\frac12 \sum_1^n \log(1-2t\lambda_j))} $$ which is the mgf for the quadratic form in the case without linear and constant term.
To use this result for the general case, write as in the question, $$ Y=X^T B X + f^t X + g $$ (where we have changed name for the constants to avoid name clashes). To use the above result we must transform $X$ to eliminate the linear term. To obtain this, replace $X$ with $X-h$ where $$ h = -\frac12 B^{-1}f $$ Then we obtain $$ Y = (X-h)^T B (X-h) +g - h^T B h $$ And then we are ready to apply the mgf found in the first part: $$ \DeclareMathOperator{\E}{\mathbb{E}} \E e^{tY} = e^{g-h^T B h} M(t) $$ where $M(t)$ is the mgf from the first part.