Two distinct elements are called "comparable" when one of them is greater than the other. This is the definition of "comparable". When you have a partially ordered set, some pairs of elements can be not comparable. i.e. you can have two elements $x$ and $y$ such that $x\leqslant y$ is false and $y \leqslant x$ is also false.
For example, consider the set $\mathbb{R}^2$ and a partial order defined like this:
$$
(x_1,x_2) \leqslant (y_1,y_2) \quad\textrm{iff}\quad x_1\leqslant y_1 \,\textrm{and}\,x_2 \leqslant y_2.
$$
With this partial order, elements $(0,0)$ and $(1,2)$ of $\mathbb{R^2}$ are comparable, because $(0,0)\leqslant(1,2)$. But elements $(0,1)$ and $(1,0)$ are not comparable, because both statements "$(0,1)\leqslant(1,0)$" and "$(1,0)\leqslant(0,1)$" are false.
The quoted material looks like part of a fairly common proof of Zorn's Lemma, so I'll try to explain it from that point of view.
Let me pretend, for a moment, that we know about ordinal numbers and the notion of definition by transfinite induction. Then there is a rather natural way to prove Zorn's Lemma, making use of a choice function $c$ on the (partially) ordered set $X$. Namely, use $c$ to select an element of $X$ (namely $c(X)$). If we're very lucky, this element is maximal in $X$ and we're done. Otherwise, there are strict upper bounds for the chosen element, and we can use $c$ to select one of these. If that's a maximal element, we're done, and otherwise we use $c$ to select a strict upper bound. Continue in this way, transfinitely, always choosing a strict upper bound for all the previously chosen elements. (One needs to verify, by transfinite induction, that the chosen elements form a chain in the usual sense, a totally ordered subset of $X$, so they always have an upper bound, by the hypothesis of Zorn's Lemma, and if this isn't maximal then there will be strict upper bounds to choose from at the next step.) Since the transfinite induction cannot continue through all the ordinals, it must end, and that happens only when it reaches a maximal element.
This argument is, in my opinion, the natural way of proving Zorn's Lemma, but it requires some preparation --- information about ordinals and transfinite induction. So some authors circumvent this preparation, by essentially proving, inside the proof of Zorn's Lemma, just enough about well-ordering and transfinite induction to make the proof work. In other words, they don't prove (or even state) any general principles of transfinite induction but just establish the one instance needed in this proof.
So their goal is to describe, without saying "transfinite induction", the same chain that I described above, using transfinite induction. The result is usually some circumlocution that doesn't explicitly say "transfinite induction" but is best understood by thinking of transfinite induction. Specifically, one ultimately wants a well-ordered chain in which each element $x$ is the chosen (by $c$) strict upper bound for all the elements that precede $x$. And the chain should continue for as long as possible, namely until a maximal element is reached (at which point there are no more strict upper bounds). This "as long as possible" is obtained by taking "approximations" that might stop short, but then forming the union of all these approximations; these approximations are the "chains" in the material quoted in the question, and the complicated definition of "chain" is just saying that each $x$ is $c$'s chosen strict upper bound for its predecessors.
The same approximation technique is used in the general proof that one can define functions by transfinite recursion, but, instead of being isolated in that general principle, it is here mixed with the rest of the argument for Zorn's Lemma, and the mixture is, in my opinion, unnecessarily confusing.
Best Answer
The text is defining a strict order, that is a relation that is irreflexive and transitive. In this case, totality is expressed by trichotomy.
You get a “standard” order relation by defining $$ x\le y \quad\text{for}\quad x<y\text{ or }x=y $$ Such a relation is indeed reflexive, antisymmetric and transitive and also a total order that satisfies
In general, strict orders and orders are essentially the same thing. A strict order on $X$ is a relation $R$ that's irreflexive ($x\mathrel{R}x$ holds for no element) and transitive.
If $R$ is a strict order on $X$, then $S=R\cup\Delta_X$ is an order on $X$, where $\Delta_X=\{(x,x):x\in X\}$. Similarly, if $S$ is an order on $X$, then $R=S\setminus\Delta_X$ is a strict order.
In the case above, $\le$ is the order associated to the strict order $<$.