You are really asking a question about invariant theory applied to the curvature tensor and its covariant derivatives. The case of scalar invariants (of any weight) was, in principle, worked out by Weyl a long time ago. I'd have to look in the literature to get the exact references (and I'm home this weekend, away from MathSciNet), but they shouldn't be hard to find. In any case, the negative weight scalars are generated by a systematic process using contractions of the curvature and its covariant derivatives with coefficients that depend on the metric $g$.

For example, in the case $n=2$, i.e., for surfaces, the story is as follows:

First, at differential order $0$, there is $g$, a nondegenerate section of $S^2(T^*)$, and its cometric $g^\sharp$, a (nondegenerate) section of $S^2(T)$. These have weight $+1$ and $-1$ respectively. Then you get all of the zeroth-order regular invariants by applying algebraic operations to these. For example, you have $\det(g)$, a nonvanishing section of $S^2\bigl(\Lambda^2(T^*)\bigr)$, which has weight $2$ and order $0$. (Note that the "area form", which would be a section of $\Lambda^2(T^*)$, is not a regular invariant, since it is not polynomial in the coefficients of $g$ in local coordinates.) Obviously, by algebra, one can generate a lot more, such as $g\otimes g$, a section of $S^2(T^*)\otimes S^2(T^*)$ and $g^{(2)}$, a section of $S^4(T^*)$, etc. However, none of these invariants (other than the trivial constant function) are regular scalar invariants of weight $0$ and order $0$.

Next, starting with the metric $g$, one can take one derivative to get the Levi-Civita connection and then another derivative to get the Gauss curvature $K = K(g)$, which has weight $-1$ (and order $2$). (You should check your conventions, because you seem to be claiming that $K$ has weight $-2$, but one obviously has $K(\lambda g) = \lambda^{-1} K(g)$.) Obviously, the $m$-th power of $K$ has weight $-m$ and order $2$, and any regular scalar invariant of weight $-m$ and order at most $2$ is a constant multiple of this $K^m$. Meanwhile an expression such as $K^2\,\det(g)$ is a regular invariant of weight $0$ and order $2$, it's just not scalar-valued.

Now, to generate higher order invariants, you do the following: Define $DK = D^1K = \mathrm{d}K$, which is a regular invariant of weight $-1$ and order $3$ that is a section of $T^* = S^1(T^*)$. Then, for $m>1$, inductively define $D^mK = \mathsf{S}(\nabla_g D^{m-1}K)$, the symmetrization of the covariant derivative $\nabla_g D^{m-1}K$, a section of $T^*\otimes S^{m-1}(T^*)$. Thus, $D^mK$ is a section of $S^m(T^*)$ that is a regular invariant of weight $-1$ and order $m{+}2$.

Then, all of the regular invariants of order at most $m{+}2$ and some fixed weight are obtained as invariant weighted homogeneous polynomial contractions starting with the following tensors $$ g, g^\sharp, K, D^1K, D^2K,\ldots, D^mK, $$ with the individual weights being $1$ for $g$ and $-1$ for all the rest.

Thus, for example, $|\nabla K|^2$ is the contraction of $g^\sharp$ with two copies of $D^1K$ using the canonical contraction mapping $S^2(T)\otimes T^*\otimes T^*\to\mathbb{R}$. It has weight $-3$ and (differential) order $3$. Meanwhile $\Delta K$, the Laplacian of $K$ is the natural contraction of $g^\sharp$ with $D^2K$, so it has weight $-2$ and order $4$.

Past this point, it becomes an algebra problem to write down generators and relations for the ring~$I_{m+2}$ of regular scalar invariants of order at most $m{+}2$. That it is a finitely generated polynomial ring that is graded by weight, follows from a theorem of Hilbert. For example, $I_0 = I_1 = \mathbb{R}$, while $I_2 = \mathbb{R}[K]$ and $I_3 = \mathbb{R}\bigl[K, |\nabla K|^2\bigr]$, etc.