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:
- $J$ is surjective (this follows as $\beta X$ is compact, $j(X)$ is dense in $K$ and $J(\beta X)$ consequently is dense in $K$ as well)
- If $J^{-1}(U)$ is open for $U\subseteq K$, then $U$ is open. This is true for any surjective map between compact Hausdorff spaces and follows from the fact that images of compact sets are compact.
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.
Like many brief and cryptic sentences found in introductions of research papers, this "concrete representation" sentence is most likely intended to be an appeal to some kind of intuition which a less fortunate reader may not share. In this case, it is an appeal to an intuitive understanding of a broad class of compactification constructions.
There are many, many, many ways to construct compactifications by embedding $X$ into compact spaces: If $f : X \to C$ is any embedding of $X$ into a compact space $C$, meaning a homeomorphism from $X$ onto a subspace of $C$, then the closure of the image $\overline{\text{image}(f)}$ may be regarded as a compactification of $X$.
Let's take $X = (0,1]$ for example.
We could embed $f : (0,1] \to \mathbb R^2$ by the formula
$$f(x) = (x,\sin(1/x))
$$
The subspace $\text{image}(f) \subset \mathbb R^2$ is bounded and is therefore contained in a closed ball, which is compact. The closure $\overline{\text{image}(f)}$ is therefore a compactification of $(0,1]$, known as the topologist's sine curve (or most of the topologist's sine curve, at least, including the most interesting portion of it). Notice that $\overline{\text{image}(f)}$ is obtained by adding a line segment to $\text{image}(f)$, namely $\{0\} \times [-1,+1]$.
Or we could choose a different embedding $f : (0,1] \to \mathbb R^2$ using the formula
$$f(x) = (r(x) \cos(\theta(x)), r(x) \sin(\theta(x)))
$$
where $r(x)= 1 - x$ and $\theta(x) = \frac{1}{x}$. Again $\text{image}(f)$ is bounded, which gives a compactification $\overline{\text{image}(f)}$ that is obtained from $\text{image}(f)$ by adding the unit circle to $\text{image}(f)$.
Now let your imagination run wild: by choosing "concrete representations" of the space $X = (0,1]$, for example representations as bounded subsets of a Euclidean space $\mathbb R^n$, we obtain many, many strange compactifications of $X$.
What that sentence means, in a rather rough and intuitive sense, is this: an $H$-compactification is not one of these random, silly, "concrete representation" compactifications.
Best Answer
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.