Since $X$ has the discrete topology, $\lim(F)$ doesn’t exist unless $F$ is a principal (fixed) ultrafilter. However, $\lim(f_*F)$ does exist, since $f_*F$ is an ultrafilter on the compact space $K$, and you do indeed want to define $\bar f(F)=\lim(f_*F)$. Let $U$ be an open nbhd of $\bar f(F)$ in $K$; then $U\in f_*F$, so $f^{-1}[U]\in F$. Let $\mathscr{U}=\{G\in\beta X:f^{-1}[U]\in G\}$; $\mathscr{U}$ is an open nbhd of $F$ in $\beta X$, and it only remains to show that $\bar f[\mathscr{U}]\subseteq U$, i.e., that if $f^{-1}[U]\in G\in\beta X$, then $\bar f(G)\in U$. But $\bar f(G)\in U$ iff $U\in f_*G$ iff $f^{-1}[U]\in G$, so this is clear.
Since you already know about the Alexandroff one-point compactification, let me begin by saying that the Stone-Cech compactification is at the other extreme, adding as many points at infinity as possible. To see what that could mean, let's consider some other compactifications of $\mathbb R$, starting with the most familiar, the extended real line, obtained by adjoining the two points $+\infty$ and $-\infty$ at the two ends of the line. Compared with the Alexandroff compactification, we're now distinguishing two different ways of "going to infinity". Some sequences that converged to $\infty$ in the Alexandroff compactification fail to converge in the extended real line because part of the sequence goes to the left and part to the right (e.g., $(-1)^nn$).
This idea can be extended, to produce "bigger" compactifications. If you visualize $\mathbb R$ as embedded in the plane as the graph of the sine function and then take its closure in the extended plane $(\mathbb R\cup\{+\infty,-\infty\})^2$, you get a compactification with a whole line segment at $+\infty$ (and another at $-\infty$). Similarly, embedding $\mathbb R$ in $3$-dimensional space as a helix, by $x\mapsto (x,\cos x,\sin x)$, we get a compactification with circles at the ends. Proceeding analogously with all bounded continuous functions $\mathbb R\to\mathbb R$ (in place of $\cos$ and $\sin$), in a very high-dimensional space (in fact, $2^{\aleph_0}$ dimensions), you get one of the standard constructions of the Stone-Cech compactification of $\mathbb R$. Roughly speaking, it separates, into different points at infinity, all of the possible "ways to go to $\infty$" in $\mathbb R$.
The analogous story works for discrete spaces $X$ in place of $\mathbb R$ (except that I don't need to say "continuous" because all functions on a discrete space are continuous). A point in the Stone-Cech remainder of a discrete space $X$ should be thought of as a "way to go to $\infty$" in $X$. But how can such "ways" be described?
Well, in any compactification, each point $p$ at infinity is in the closure of the original space $X$, and so we can describe its location relative to $X$ by the trace on $X$ of its neighborhood filter, i.e. by $\mathcal F=\{U\cap X: p\in\text{interior}(U)\}$ (where $U$ refers to subsets of the compactification). Note that $\mathcal F$ can't contain any finite subsets of $X$, because such subsets are closed in the compactification (as in any $T_1$ space) and thus disjoint from suitable neighborhoods of any point $p$ at infinity.
For the Stone-Cech compactification of a discrete space $X$, this filter $\mathcal F$ must have one additional property, namely that we cannot have two disjoint subsets $A,B$ of $X$ both meeting all the sets in $\mathcal F$. The reason is that then "going to infinity" in $A$ and in $B$ would be two different ways to go to infinity, both leading to the same point $p$.
This additional property of the filter $\mathcal F$ is equivalent to saying that $\mathcal F$ is an ultrafilter. So this is how ultrafilters enter the picture of Stone-Cech compactifications of discrete spaces.
One then ordinarily continues by saying that, (1) since the filter $\mathcal F$ associated with any $p$ at infinity is an ultrafilter, and different $p$'s must correspond to different ultrafilters (in order for the compactification to be Hausdorff), we might as well identify the points $p$ with the ultrafilters $\mathcal F$, and (2) for the sake of uniformity, we might as well identify the principal ultrafilters (which haven't been used yet) with the points of $X$. With these conventions, one can prove (using compactness) that every ultrafilter on $X$ gets identified with a point in the Stone-Cech compactification of $X$. So the Stone-Cech compactification of a discrete space can be identified with the set of ultrafilters on $X$. Finally, one should verify that the topology is necessarily the one you described.
Best Answer
Yes. Indeed, let $U\subseteq A$ be an open subset. Then $h^{-1}(U)$ is the set of ultrafilters on $A$ whose limit is in $U$. Now if the limit $h(F)$ of an ultrafilter $F$ is in $U$, then there is a neighborhood $V$ of $h(F)$ whose closure is contained in $U$. Then $V\in F$, and moreover every ultrafilter which contains $V$ has limit in $\overline{V}\subseteq U$. Thus the basic open set $[V]\subseteq\beta(A)$ is a neighborhood of $F$ which is contained in $h^{-1}(U)$. Since $F\in h^{-1}(U)$ was arbitrary, $h^{-1}(U)$ is open. So the inverse image of an open set under $h$ is open, and so $h$ is continuous.
Note also that this is literally the content of the associative property of an algebra over the ultrafilter monad (so, if you already know that $\beta$-algebras are the same as compact Hausdorff spaces you're done, and conversely this is a key step in proving that $\beta$-algebras are the same as compact Hausdorff space). The associative property for the structure map $h:\beta(A)\to A$ says that the the diagram $$\require{AMScd} \begin{CD} \beta(\beta(A)) @>{\beta(h)}>> \beta(A)\\ @V{\mu_{\beta(A)}}VV @V{h}VV \\ \beta(A) @>{h}>> A \end{CD}$$ commutes. The composition $h\beta(h)$ takes an ultrafilter on $\beta(A)$, pushes it forward to $\beta(A)$ along $h$, and then takes its limit in $A$. The composition $h\mu_{\beta(A)}$ takes an ultrafilter on $\beta(A)$, computes its limit in the topology of $\beta(A)$ (that's what $\mu$ does), and then takes the image under $h$. So to say that these two maps are equal is to exactly say that the map $h$ preserves convergence of ultrafilters, with respect to the topologies on $\beta(A)$ and $A$. That is, $h$ is continuous.
In fact, this is similarly true for any monad. If $T$ is a monad on some category and $h:T(A)\to A$ is a $T$-algebra, then $T(A)$ has a canonical $T$-algebra structure via $\mu_{T(A)}$ and the associativity axiom for $h$ says exactly that $h$ is a $T$-algebra homomorphism from $T(A)$ to $A$.