The sentence you wrote in your OP makes no sense.

The sentence which you intend to quote reads:

[T]he metric $g_{v_\infty}$, on $M\setminus \{x_\infty\}$ has an 'asymptotically flat end at $x_\infty$'.

Now, $g_{v_\infty}$ is **not** a Riemannian metric on $M$. The metric $g_{v_\infty}$ is constructed from $g$ with a conformal weight given by $v_\infty$, which by definition only exists on $M\setminus\{x_\infty\}$ as the function $v_\infty$ blows up at $x_\infty$. Hence $g_{v_\infty}$ is a Riemannian metric only on $M\setminus \{x_\infty\}$.

*Therefore*, the correct interpretation of the sentence you quoted is that

- The (non-compact) Riemannian manifold $(M\setminus \{x_\infty\}, g_{v_\infty})$ is asymptotically flat
- It only has one end, which is a neighborhood of $x_\infty$ in the following sense. There exists a neighborhood $V$ of $x_\infty$ (in the topology of the smooth compact manifold $M$) such that $M\setminus V$ is compact (automatically true) and that there exists a diffeomorphism of $V\setminus \{x\}$ with $\mathbb{R}^d \setminus \overline{B(1,0)}$ such that the usual asymptotic flatness decay condition for the metric $g_{v_\infty}$ holds.

In other words, the point $x_\infty$ is "the point at infinity" for its corresponding asymptotic flat end, or that one can regard $(M,g)$ as a one-point conformal compactification of $(M\setminus \{x_\infty\},g_{v_\infty})$