Item lookaheads versus dot lookaheads for $LR(k)$ with $k \gt 1$?
I think you are mistaken, they are needed but the dot look-ahead there is so obvious that you have not paid attention to the fact it is used.

First, let's remark that there are three kinds of items:


those in which the dot is just before a non-terminal. They never participate in an ambiguous situation: when a non-terminal has been produced, it is shifted.
those in which the dot is at the end. They have the item look-ahead and the dot look-ahead which are equal (what may follow the dot is what may follow the produced non-terminal when the production is reduced as the dot is at the end of the item).
those in which the dot is just before a terminal.  They have the item look-ahead and the dot look-ahead which are different.  The item look-ahead is what may follow the non-terminal when the production is reduced, the dot look-ahead start by the terminal which follow the dot and continue with what can be generated after that terminal.


Now, with a look-ahead of 1 or less, the dot look-ahead is trivial: either it is the item look-ahead or the terminal which is just after the dot and that's what you are using to solve a conflict (or decide that there is no way with the limited look-ahead you have).  With a look-ahead of 2 or more, you have to compute the dot look-head or you may not know if you have to shift or to reduce as in the example provided by Grune and Jacobs:

$$\begin{array}{l}
S \rightarrow Aa \; | \; Bb \;| \;Cec \;|\; Ded \\
A \rightarrow qE \\
B \rightarrow qE \\
C \rightarrow q \\
D \rightarrow q \\
E \rightarrow e \\
\end{array}$$

which has the state:
$$\begin{array}{lcc}
&amp;\textrm{item look-ahead}&amp;\textrm{dot look-ahead}\\
A \rightarrow q \cdot E &amp; a\#  &amp; ea\\
B \rightarrow q \cdot E &amp; b \# &amp; eb\\
C \rightarrow q \cdot   &amp; ec   &amp; ec\\
D \rightarrow q \cdot   &amp; ed   &amp; ed\\
E \rightarrow \cdot e   &amp; a \# &amp; ea\\
E \rightarrow \cdot e   &amp; b \# &amp; eb\\
\end{array}$$