I think the implication you are asking about might not in fact hold.

If my computations are correct, we can see this already for $S_4$.

Let me ignore the dot action part of the question (I don't think that matters) and try to simplify it as follows: let $\lambda$ be any regular anti-dominant weight, and define a partial order $\preceq$ on $W$ by $u \preceq v$ if and only if $u\lambda \leq v\lambda$ (this is the usual partial order on weights with $\mu \leq \lambda$ if and only if $\lambda-\mu=\sum_{i}c_i\alpha_i$ with $c_i\in\mathbb{Z}_{\geq 0}$). The question becomes: is $\preceq$ the same partial order as $\leq$, the usual (strong) Bruhat order?

My answer is that, no, they are not the same and this can be seen already for $S_4$. For instance, let us take $\lambda =(1,2,3,4)$ as our regular anti-dominant weight so that $w\lambda$ is just the one-line notation of the permutation $w \in S_4$. Then I claim that $(1,4,2,3) \preceq (2,3,4,1)$ but $(1,4,2,3)\not \leq (2,3,4,1)$. That $(2,3,4,1)-(1,4,2,3)= 1*(1,-1,0,0) + 2*(0,0,1,-1)$ shows $(1,4,2,3) \preceq (2,3,4,1)$. And my computer tells me that $(1,4,2,3)\not \leq (2,3,4,1)$ (actually I think that this is also easy to see since we cannot ever move the $4$ rightward in $(1,4,2,3)$ when going up in Bruhat order).

EDIT:

Regarding the dot action aspect of the question, here is a larger portion of the text in question which shows the notions of regular and antidominant are meant relative to the dot action:

and$w < s w$ (although I'm still not sure how to do that). $\endgroup$5more comments