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.
Question 1: No.
Let $X = [0,1] \setminus N$ with $N = \{1/n \mid n \in \mathbb N\}$. This space is not locally compact because $0$ does not have any compact neigborhood. It has $[0,1]$ as a compactification, but here it has $N$ as a remainder which is not dense.
Question 2: Yes.
Daniel Wainfleet has given an example: Take $X = [0,1] \cap \mathbb Q$. It has $[0,1]$ as a compactification different from Stone-Čech compactification, and the remainder is dense.
Best Answer
If $K$ is any compactification of $X$ (by which I mean a compact Hausdorff space with $j:X\to K$ a dense embedding), then the definition of the Stone-Cech-compactification provides you with a continuous map $J:\beta X\to K$ such that $J\circ i = j$, where $i:X\to\beta X$ is the embedding of $X$ into its Stone-Cech-compactification. This is the crucial aspect to know about the Stone-Cech-compactification: Any other map from X into a compact Hausdorff space factors through $\beta X$. Now $J$ is a continuous map between two compact Hausdorff spaces. To see that it is a quotient map you just need to prove the following:
I don't quite understand your remark where you want to consider the one-point-compactification only. This is just a special case that generalises as sketched above (usually there are many different compactifications of a space, not just the two mentioned so far), but is not sufficient in itself.