Is this modification of connexity necessary, or redundant in the definition of partial ordering?

It is necessary that in the definition of ordering in the sense of $<$, the connex property is replaced by the condition $\forall x \forall y (x \neq y \implies (x\,R\,y \lor y\, R\, x))$. Otherwise, there would not exist any ordering in the sense of $<$, because the connex property and the irreflexive property (i.e. $\forall x \, \lnot(x\,R\,x)$) are incompatible: if one of the two holds, the other is false.

Indeed, suppose that an ordering in the sense of $<$ is defined as a relation $R$ that is

1. transitive: $\forall x \forall y \forall z ((x\,R\,y \land y\,R\,z) \implies x\,R\,z)$
2. irreflexive: $\forall x \, \lnot (x\,R\,x)$
3. connex: $\forall x \forall y (x\,R\,y \lor y\, R\, x)$.

Proposition. In every non-empty set, there is no ordering in the sense of $<$.

Proof. By contradiction, suppose that $R$ is an ordering in the sense of $<$ on a non-empty set $S$. Let $z \in S$ (it exists because $S$ is not empty). By the connex property taking $x = y =z$, we have $z\,R\,z \lor z\,R\,z$, that is, $z\,R\,z$. But this is impossible because of the irreflexive property.$\qquad\square$

What is wrong in your reasoning? You claim that, because of irreflexivity, we already know that $x \neq y$ and so this hypothesis $x \neq y$ in the condition $\forall x \forall y (x \neq y \implies (x\,R\,y \lor y\, R\, x))$ is superfluous and can be dropped, since it always holds.

Wait! The convex property says that for every $x$ and $y$ in the set $S$ where the relation $R$ is defined, we have $x\,R\,y$ or $y\, R\, x$. By irreflexivity, this implies that $x \neq y$. But convexity is talking about every $x$ and $y$ in the set $S$, also when $x = y$. Summing up, irreflexivity and convexity together are saying that whenever you take whatsoever $x$ and $y$ in $S$, necessarily we have $x \neq y$; but this is absurd, because you can always choose $x = y$ in $S$.

To avoid this contradiction, in the definition of ordering in the sense of $<$, the hypothesis $x \neq y$ added to the convex condition is not redundant, but necessary.

Said differently, you are right when you say that if you assume irreflexivity and convexity, then

\begin{align}\tag{*} \text{for every $x$ and $y$ we have $x \neq y$} \end{align} Your error is that you did not notice that $(*)$ is claiming something impossible, and hence irreflexivity and convexity cannot coexist: there is no relation that is both irreflexive and connex.

A terminological remark. The terminology used in your reference is quite unusual. Often, the "ordering in the sense of $\leq$" is called total or linear (nonstrict) order, the "ordering in the sense of $<$" is called total (or linear) strict order, the "connex" property is called total or linear or strong connnected, the "identitive" property is called antisymmetric.