The following is an excerpt from Page 10 of this paper by Fefferman-Graham.
We prepare to define ambient spaces in the even-dimensional case. Let $S_{IJ}$ be a symmetric 2-tensor field on an open neighborhood of $G\times \{0\}$ in $G\times \Bbb{R}$. For $m\geq 0$, we write $S_{IJ}=O^+_{IJ}(\rho^m)$ if
i) $S_{IJ}=O(\rho^m)$
What does $O(\rho^m)$ mean? I would have guessed that it has something to do with order. However, I couldn't find a reference to $\rho$ before. So I really don't know what they mean by it in this case.
As mentioned at the beginning of the first full paragraph on the cited page, $\rho$ is the standard coordinate on the factor $\Bbb R$ in the product $\mathcal G \times \Bbb R$. So, if one fixes coordinates $(y^0, \ldots, y^n)$ on $\mathcal G$, then $(y^0, \ldots, y^n, \rho)$ are coordinates on $\mathcal G \times \Bbb R$. Then, $\color{blue}{S_{IJ} = O(\rho^m)}$ is just the condition that each of the component functions of $S$ with respect to these coordinates is $O(\rho^m)$.
Explicitly, we can expand any smooth function $f$ on $\mathcal{G} \times \Bbb R$ in a Taylor series in $\rho$: $$f(y^0, \ldots, y^n, \rho) \sim f_0(y^0, \ldots, y^n) + f_1(y^0, \ldots, y^n) \rho + f_2(y^0, \ldots, y^n) \rho^2 + \cdots,$$ where we can view the coefficients $f_i$ as functions on $\mathcal G$. Then the condition that $f$ is $O(\rho^m)$ is just that $f_0 = f_1 = \cdots = f_{m - 1} = 0$.
It's routine to check that the definition of $O(\rho^m)$ is independent of the choice $(y^0, \ldots, y^n)$ of coordinates, so the definition is geometric
Remark As mentioned on the previous page in the article, one way to get coordinates on $\mathcal G$ is to fix a representative metric in the given conformal class and pick a metric on the manifold $M$ underlying $\mathcal G$ and coordinates $(x^a)$ on $M$, giving coordinates $(t, x^a)$ on $\mathcal G$ and thus coordinates $(t, x^a, \rho)$ on $\mathcal G \times \Bbb R$---these latter coordinates are those used in the proofs in Section 3 of the main results.