one-point compactification of a compact space

If we take your definition ($Y$ compact Hausdorff and $y_0 \in Y$ such that $i: X \rightarrow Y \setminus \{y_0\}$ is a homeomorphism), the answer is clear:

$i[X] = Y \setminus \{y_0\} \subset Y$ is compact and hence closed in $Y$ (as $Y$ is Hausdorff), and so $\{y_0\}$ is open (i.e. $y_0$ is an isolated point of $Y$). So $Y$ then consists of a (closed) topological copy of $X$ (namely $Y \setminus \{y_0\}$) and an isolated point $y_0$. Indeed $i[X]$ is not dense in $Y$. So indeed this is incompatible with the usual definition of a compactification:

Commonly, a (Hausdorff) compactification of a space $X$ is a pair $(Y, i)$ where $Y$ is compact Hausdorff and $i : X \rightarrow Y$ is an embedding (or equivalently, $i$ is a homeomorphism between $X$ and $i[X] \subset Y$) and $i[X]$ is dense in $Y$. In this definition, a one-point compactification $X$ is a Hausdorff compactification $(Y,i)$ such that moreover $Y \setminus i[X]$ is a singleton.

Two compactications $(Y,i)$ and $(Y',i')$ of $X$ are called equivalent when there is a homeomorphism $h: Y \rightarrow Y'$ such that $h \circ i = i'$ as maps on $X$. One then shows that $X$ has a one-point compactification iff $X$ is locally compact Hausdorff and non-compact and moreover all of them are equivalent in the above sense.

In this general definition, if $X$ is compact and $(Y,i)$ is a Hausdorff compactification, $i[X]$ is compact and so closed, so it can only be dense in $Y$, $i[X] = Y$ and $Y$ is just a homeomorph of $X$. So there is no one-point Hausdorff compactification, in the general sense, when $X$ is already compact.

So maybe your text does not use the more common definition that includes denseness, or it implicitly assumes all considered spaces are non-compact, in which case we have no problem.

If $X$ is compact already, the one point compactification is the disjoint union of $X$ and a point $y_0$.

To understand the one point compacification, you only need to know what the neighborhoods of $y_0$ are. But they are by definition given by $\{y_0\}\cup (X-K)$ where $K\subset X$ is compact. Hence $\{y_0\}$ is actually open when $X$ is compact.