Just to agree on notation: A space is zero-dimensional if it is $T_1$ and has a basis consisting of clopen sets, and totally disconnected if the quasicomponents of all points (intersections of all clopen neighborhoods) are singletons. A space is hereditarily disconnected if no subspace is connected, i.e., if the components of all points
are singletons. (Edit: There seems to be disagreement about the names of these properties.
Often what I call hereditarily disconnected is called totally disconnected and what I call totally disconnected is then called totally separated.)
Note that zero-dimensionality implies Hausdorffness.
Zero-dimensional implies totally disconnected since every point can be separated from every other point by a clopen set.
Totally disconnected implies hereditarily disconnected: given a set $A$ with at least two points, one point is not in the quasi-component of the other and hence the two points can be separated by a clopen set. Hence the set $A$ is not connected.
This shows that the space is hereditarily disconnected.
On the other hand, if $X$ is locally compact and hereditarily disconnected, take $x\in X$
and let $U$ be an open set containing $x$.
By local compactness, find an open neighborhood $V\subseteq U$ of $x$ whose closure $\overline V$ is compact.
In a compact space, components and quasi-components coincide and hence the quasi-component of $x$ in $\overline V$ is $\{x\}$ (you don't need this if you are not interested in hereditary disconnected spaces but just totally disconnected ones). Using compactness again,
there are finitely many clopen subsets $C_1\dots,C_n$ of $\overline V$ such that
$x\in C_1\cap\dots\cap C_n\subseteq V$. The intersection of the $C_i$ is closed in $X$ since this intersection is compact.
It is open in $X$ since it is open in $V$ and $V$ is open in $X$.
This shows that the clopen subsets of $X$ form a basis.
Hence $X$ is zero-dimensional.
Edit: As suggested by Joseph Van Name, I include a proof that in a compact space the components coincide with the quasi-components.
Let $X$ be a compact space and $x\in X$. The component $C$
of $x$ is the union of all connected subsets of $X$ containing $x$. If $A\subseteq X$ is clopen and $x\in A$, then the component of $x$ is contained in $A$. It follows that the component of $x$ is contained in the quasi-component $Q$ of $x$.
In order to show that the component $C$ and the quasi-component $Q$ coincide,
it is now enough to show that $Q$ is connected.
Observe that $Q$ is closed in $X$ and thus compact.
Now suppose that $Q$ is not connected. Then there are nonempty,
relatively open subsets $A$ and $B$
of $Q$ such that $A\cap B=\emptyset$ and $A\cup B=Q$.
Note that $A$ and $B$ are relatively closed in $Q$ and hence compact.
Hence $A$ and $B$ are closed in $X$.
Two disjoint closed sets in a compact space can be separated by open subsets, i.e.,
there are disjoint open sets $U,V\subseteq X$ such that $A\subseteq U$ and $B\subseteq V$.
We have $$Q=\bigcap\{F\subseteq X:F\mbox{ is clopen and }x\in F\}$$
and thus
$$\bigcap\{F\subseteq X:F\mbox{ is clopen and }x\in F\}\cap(X\setminus(U\cup V))=\emptyset.$$
By compactness there are finitely many clopen sets $F_1,\dots,F_n$ containing $x$ such that
$$F_1\cap\dots\cap F_n\cap(X\setminus(U\cup V))=\emptyset.$$
Let $F=F_1\cap\dots\cap F_n$.
$F$ is clopen and we have $Q\subseteq F\subseteq U\cup V$.
We have
$$\overline{U\cap F}\subseteq\overline U\cap F=\overline U\cap(U\cup V)\cap F=U\cap F.$$
It follows that $U\cap F$ is clopen in $X$.
We may assume $x\in A$.
Since $B$ is nonempty, there is some $y\in B$.
But now $y\not\in U\cap F$. It follows that $y$ is not in the quasi-component of $x$,
a contradiction.
As you refer to Engelking's "Dimension theory" book, I suppose you know the following two statements, but anyway, here they are:
The notions agree for separable metric spaces, by Exercise 1.7.E of Engelking's book.
The notions agree for paracompact spaces by Proposition 3.2.2 of Engelking's book. (In particular, if both dimensions are finite, they agree.)
An example where the two notions differ was suggested in the comment of Benoît Kloeckner: the long ray $[0,\omega_1)\times[0,1)$ with the lexicographic order topology. The long ray is locally compact but not paracompact. If we use the definition of covering dimension with arbitrary open covers, the long ray has infinite covering dimension (because finite covering dimension implies paracompactness). However, for the finitary covering dimension we get 1 as follows. Any finite open cover is essentially a couple of initial segments $((\lambda_1,a_i),(\lambda_2,b_i))$ with $\lambda_i$ countable ordinals and at least one long end $((\lambda,a),(\omega_1,1))$. The segments with the countable ordinals in the upper bound are homeomorphic to intervals $(a,b)\subseteq\mathbb{R}$. For the initial segments, we get a refinement such that at most two-fold intersections are nonempty (as one would do in $\mathbb{R}$), and only the last of the initial segments intersects the long end $((\lambda,a),(\omega_1,1))$.
Addendum: I just noted that $[0,\omega_1)$ with the order topology is an even simpler (and slightly more puzzling) example. It is locally compact, Hausdorff and not paracompact, but its finitary covering dimension is $0$.
Best Answer
You can read a review of the paper in Zentralblatt, it contains a short description in German.
The review on MathSciNet is a bit more extensive (but requires a subscription). There is indeed the condition of having an $n+1$-to-one map from a zero-dimensional compact space onto the space itself. There is also a condition on `gratings' (not defined, but my guess is, based on other papers: a grating is a finite closed cover where the interiors of the closed sets are pairwise disjoint).
In What is a non-metrizable analog of metrizable compacta? (Part I) Pasynkov defines a class of compacta where the dimensions coincide.