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It is easy to find references stating that the category of compact Hausdorff spaces $mathbfCompHaus$ is equivalent to the category of algebras for the ultrafilter monad, $mathbfbeta Alg$. After doing some digging, the $mathbfCompHausto mathbfbeta Alg$ half of the equivalence is simple enough, but I haven't been able to find a description of the $mathbfbeta Algto mathbfCompHaus$ half of this equivalence. I have tried to work it out, but I have little experience with topological spaces and am not sure what the associated topology ought to look like.

I'm wondering if anyone has a good reference that describes the $mathbfbeta Algto mathbfCompHaus$ half of the equivalence, or can describe it here.

The other half of the equivalence is described on the nLab page on ultrafilters.

Given an algebra structure $xicolon beta X to X$, we define the topology on $X$ by declaring that a subset $Usubseteq X$ is open if and only if for every point $xin U$ and every ultrafilter $Fin beta X$ such that $xi(F) = x$, we have $Uin F$.

The idea is simple: in a topological space $X$, a subset $Asubseteq X$ is closed iff it is closed under limits of ultrafilters. That is, $A$ is closed iff for any ultrafilter $F$ supported on $A$ (i.e., $Ain F$), all limits of $F$ are in $A$.

Now suppose we are given a $beta$-algebra $L:beta Xto X$. The idea is that $L$ takes each ultrafilter to its limit. So, a subset $Asubseteq X$ should be closed iff it is closed under limits of ultrafilters according to $L$: that iff $L(F)in A$ for all $Finbeta X$ such that $Ain F$. Or, restating this contrapositively in terms of open sets, a set $U$ is open iff for all $Finbeta X$, $L(F)in U$ implies $Uin F$.

It is easy to verify that this does define a topology on $X$, and that every ultrafilter $F$ converges to $L(F)$ with respect to this topology (it is essentially by definition the finest topology such that this is true). Verifying that $L(F)$ is the unique limit of $F$ (and therefore the topology is compact Hausdorff since each ultrafilter has a unique limit) is trickier and requires you to use the associativity and unit properties of $L$ as a $beta$-algebra.

In detail, first let $Asubseteq X$ and define $$C(A)=L(F):Finbeta X,Ain F.$$ I claim that $C(A)$ is closed. Suppose $Finbeta X$ and $C(A)in F$. Let $mathcalF$ be the filter on $beta X$ generated by the sets $L^-1(B)$ for $Bin F$ together with $Ginbeta X:Ain G$; these have the finite intersection property since every $Bin F$ has nonempty intersection with $C(A)$. Let $mathcalG$ be an ultrafilter on $beta X$ extending $mathcalF$ (so $mathcalGinbetabeta X$). We now use the associativity property of $L$, which says that $$L(lim mathcalG)=L(beta L(mathcalG))$$ where $lim:betabeta Xto beta X$ is the structure map of the monad $beta$ at $X$. Explicitly, $limmathcalG$ is defined as the set of $Bsubseteq X$ such that $Ginbeta X:Bin Gin mathcalG$. In particular, by our choice of $mathcalG$, we have $AinlimmathcalG$. On the other hand, $beta L(mathcalG)$ is by definition the set of $Bsubseteq X$ such that $L^-1(B)inmathcalG$, and so $Fsubseteq beta L(mathcalG)$. Since $F$ is an ultrafilter, this means $beta L(mathcalG)=F$. We thus conclude that $$L(limmathcalG)=L(F)$$ where $AinlimmathcalG$, and thus $L(F)in C(A)$.

So, for any $Asubseteq X$, $C(A)$ is closed. Note also that $Asubseteq C(A)$ by the unit property of $beta$: $L$ sends each principal ultrafilter to the corresponding point. (It follows easily that in fact $C(A)$ is the closure of $A$, though we will not use this.)

Now let $Finbeta X$ and $xin X$ and suppose $F$ converges to $x$ in our topology; we will show that $x=L(F)$. Since $F$ converges to $x$, every closed set in $F$ contains $x$. Thus for all $Ain F$, $xin C(A)$, since $C(A)$ is a closed set and $C(A)in F$ since $Asubseteq C(A)$. Let $mathcalF$ be the filter on $beta X$ generated by the sets $T_A=Ginbeta X:L(G)=x,Ain G$ for $Ain F$; they have the finite intersection property since $T_Acap T_B=T_Acap B$ and $xin C(A)$ for all $Ain F$. Extend $mathcalF$ to an ultrafilter $mathcalG$ on $beta X$. Similar to the argument above, we then have $F=limmathcalG$ and $xin beta L(mathcalG)$ so $beta L(mathcalG)$ is the principal ultrafilter at $x$. By the associativity and unit properties of $L$, we thus have $$L(F)=L(limmathcalG)=L(beta L(mathcalG))=x.$$

Morally, what is going on in these arguments is that the associativity property of $L$ says that $L$ is "continuous" from the standard topology on $beta X$ to $X$ (see this answer of mine for some elaboration on that idea). So, to prove that $C(A)$ is closed, if we have a net in $C(A)$ converging to a point $x$, we can choose ultrafilters supported on $A$ which $L$ maps to each point of this net. Then if we take an appropriate accumulation point of these ultrafilters as points of $beta X$, $L$ will map this accumulation point to $x$ by continuity. This will then give an ultrafilter which is supported on $A$ (because it is a limit of ultrafilters supported on $A$) which $L$ maps to $x$, to prove that $xin C(A)$.

Then, if $F$ converges to $x$, we have $xin C(A)$ for all $Ain F$, so we can pick ultrafilters $G$ with $L(G)=x$ which are supported on arbitrarily small elements of $F$. These ultrafilters then converge in $beta X$ too $F$, and so continuity of $L$ says $L(F)=x$ as well.