Often (like in the case of the standard compactifications of $\mathbb R^{d,1}$ or $\mathrm{AdS}_{d+1}$), the non-compact manifold $M$ that we start with is mapped to the interior of a compact manifold $\tilde M$ that also happens to be a manifold with boundary. In these cases, we define the conformal boundary as $\partial\tilde M$ as indicated by Ben Crowell.
Note, however, that this does not always happen. Consider, for example, the standard stereographic projection which maps $M=\mathbb R^2$ onto $\tilde M=S^2\setminus \{(0,0,1)\}$ where in my notation I'm treating the sphere as an embedded submanifold of $\mathbb R^3$ with north pole $(0,0,1)$. Notice, in this case, that $\tilde M$ is not the interior of a compact manifold with boundary; when we include the north pole, we get $S^2$ which is a compact manifold without boundary.
In contrast, consider $\mathbb R^{d,1}$ with metric
$$
ds^2 = -dt^2 + dr^2 + r^2 d\Omega_{d-1}^2
$$
where $d\Omega_{d-1}^2$ is the metric on $S^{d-1}$. Let $\vec\theta$ be coordinates on the sphere, then the diffeomorphism
$$
f(t,r,\vec\theta) = (T(t,r,\vec\theta), R(t,r,\vec\theta), \vec\theta)
$$
where
\begin{align}
T(r,t,\vec\theta) &= \tan^{-1}(t+r) + \tan^{-1}(t-r)\\
R(r,t,\vec\theta) &= \tan^{-1}(t+r) -\tan^{-1}(t-r)
\end{align}
leaves the sphere factor unchanged but maps all of the $(r,t)$ plane to the interior of the triangular region in the $(R,T)$ plane satisfying
$$
0\leq R \leq \pi, \qquad |T|\leq \pi-R
$$
This region does have a boundary (the edges of the triangle) which allows us to define the conformal boundary of $\mathbb R^{d,1}$.
As for the $\Omega^2 = 0$ constraint, this is how I think of it intuitively (and pretty imprecisely). The new metric $\tilde g$ on the compact manifold is related to the original metric $g$ by the conformal factor:
$$
\tilde g = \Omega^2 g
$$
Now in the original manifold, as you go out to infinity, $g$ allows for distances between points to be arbitrarily large. But after the compactification, all points will be some finite distance away from one another. In order for this to happen, distances between points need to be multiplied by a smaller and smaller number as you go further and further out so that the product remains finite. The factor $\Omega^2$ multiplying $g$ is precisely what does this for you. This roughly is what the quote means when it says
This reflects the property of a conformal compactification that “brings infinity to a finite distance."
By the way, I found it useful to explicitly go through the $\mathrm{AdS}_{d+1}$ example. In particular, you can, for example, verify for yourself that for the explicit mapping written above, the conformal factor is
$$
\Omega(t,r,\vec\theta)^2 = \frac{1}{\frac{1}{4}(1+(r-t)^2)(1+(r+t)^2)}
$$
which vanishes as $r,t\to\infty$
