Assume that $g$ and $g'$ are metric tensors with one dimensional kernel and the same signature.


Does the classical results of Weyl (dim >3) or of Cotton (dim=3) generalise to that case, i.e. $g$ and $g'$ are conformally equivalent iff their Weyl (resp. Cotton) tensors are equal?



Conformal equivalence degenerate metric tensors
In order to define an actual Weyl or Cotton tensor, one has to have a non-degenerate conformal structure. In the OP's case, we aren't given such a structure, so the only way to get an actual Weyl or Cotton tensor would be to construct a conformal structure out of the given data, and that, as Ben McKay points out, is not always possible.


Meanwhile, specifying, up to a multiple, a smooth semi-definite quadratic form with nullity 1 on a manifold $M$ of dimension $n$ is specifying a section of a smooth bundle over $M$ of rank $\tfrac12(n^2+n-4)$ which is greater than $n$ when $n>2$, so there must be local invariants, i.e., tensor fields that can be used to distinguish different sections up to local diffeomorphism. However, the order of the lowest order invariant tensor fields turns out to be lower than either the tensors of Weyl (order 2 for $n>3$) or Cotton (order 3 for $n=3$).


For example, consider the case of a semidefinite quadratic form $g$ of rank 2 on a $3$-manifold, defined up to a multiple. In this case, there is a tensor of order $1$ that defines a quadratic form on the null curves of the 'degenerate conformal structure' $[g]$. This tensor vanishes if and only if the degenerate conformal structure is locally the pullback of a conformal structure on the (locally defined) surface that is the space of null curves of $[g]$. Intuitively, you can understand this as follows: Since the construction is local, one can assume that one is working in a neighborhood of the origin in $xyz$ space and that the $g$-null curves are parallel to the $z$-axis. Thus,
$$
[g]= \bigl[E(x,y,z)\,\mathrm{d}x^2 + 2F(x,y,z)\,\mathrm{d}x\,\mathrm{d}y + G(x,y,z)\,\mathrm{d}y^2\bigr]
$$
where $EG-F^2>0$. Then (if I haven't made an algebra error), the quadratic form
$$
\zeta = \frac{(E\_zG{-}EG\_z)^2 - 4(E\_zF{-}EF\_z)(F\_zG{-}FG\_z)}{(EG-F^2)^2}\,\mathrm{d}z^2
$$
pulls back to each $g$-null curve to define a non-negative quadratic form on that curve that is independent of the choice of coordinates. If $\zeta$ vanishes identically, then, up to scale, one can choose the local coordinates so that $[g] = [\mathrm{d}x^2+ \mathrm{d}y^2]$, and conversely.


Meanwhile, when $\zeta$ is nowhere vanishing, using one more derivative and a choice of orientation of the $g$-null curves, define a scalar differential invariant $C$ of order 2, namely
$$
C = \frac{(EG-F^2)^{3/2}\,\det\pmatrix{E & F & G\\E\_z & F\_z & G\_z\\E\_{zz} & F\_{zz} & G\_{zz}}}{\bigl((E\_zG{-}EG\_z)^2 - 4(E\_zF{-}EF\_z)(F\_zG{-}FG\_z)\bigr)^{3/2}}.
$$  

(Reversing the orientation on the $g$-null curves, reverses the sign of $C$. Thus, $C^2$ does not depend on the choice of orientation.)
In case, $C$ is constant, it can be shown that there are local coordinates $(x,y,z)$ centered around any given point in which $[g]$ has the form
$$
[g] = \bigl[E(z)\,\mathrm{d}x^2 + 2F(z)\,\mathrm{d}x\,\mathrm{d}y + G(z)\,\mathrm{d}y^2\bigr],
$$
where $EG-F^2 = 1$, $\zeta = \mathrm{d}z^2$, and $eE-2fF+gG = h$,
where $e,f,g,h$ are constants. (For example, when $C=0$, the linear relation can be taken to be $F=0$.) Hence, the cases (with $\zeta$ nonvanishing) with $C$ constant are all locally homogeneous.


When $C$ is not constant, $(M,[g])$ is not locally homogeneous,
and the further construction of geometric invariants divides into the cases where $C\_z$ vanishes and where it does not vanish.


Using Cartan's Equivalence Method, which is more systematic than just guessing formulae, one can derive a complete set of tensorial invariants for the geometric structure $[g]$.


Similar ideas work in all dimensions $n\ge 3$.


