Gauss' theorem for null boundaries
Gauss' Theorem has nothing to do with the (pseudo-)metric. Is just a consequence of Stokes' theorem.
Stokes's theorem says that, for any $n-1$ form $\omega$,
$$ \int_M d\omega = \int_{\partial M} \omega. $$
Now fix any smooth measure $\mu$ (i.e. given by a smooth non-vanishing top dimensional form, or a density if $M$ is not orientable). It might be the Riemannian measure of a Riemannian structure, but it's not relevant.
Let $X$ be a vector field. If you apply the above formula to the $n-1$ form $\omega:=\iota_X \mu$, you get, using one of the many definitions of divergence (associated with the measure $\mu$)
$$\int_M \mathrm{div}(X) \mu = \int_{\partial M} \iota_X \mu, \qquad (\star),$$
where $\iota_X$ is the contraction. This holds whatever is $X$, on any smooth manifold with boundary.
Now it all boils down to how you want to define a "reference" measure on $\partial M$. As you propose, you can pick a transverse vector $N$ to $\partial M$, and define a reference measure on $\partial M$ as $\eta:=\iota_T \mu$. Then
$$ \int_M \mathrm{div}(X) \mu = \int_{\partial M} f\, \iota_T \mu, $$
where
$$f := \frac{\mu(X,Y_1,\ldots,Y_{n-1})}{\mu(T,Y_1,\ldots,Y_{n-1})}. $$
for any arbitrary local frame $Y_1,\ldots,Y_{n-1}$ tangent to $\partial M$ (oriented, otherwise take densities and absolute values). In particular, you can easily compute $f$ by looking at the "$T$" component of "$X$":
$$X = f\, T \mod \mathrm{span}\{Y_1,\ldots,Y_{n-1}\}.$$
There is no need to use orthonormality, or a (pseudo)-metric.
However, if you are on a Riemannian manifold, $\mu = \mu_g$ is the Riemannian measure, and $N$ is the normal vector to $\partial M$, then $f= g(N,X)$, and $\iota_N \mu$ is Riemannian measure of the induced Riemannian metric on $\partial M$, recovering the classical statement.