Thanks for your patience with my reply.

Short Answer:

Yes, it should be $d-c+b-a$ to be consistent and it is now updated.

Long Answer (+ preview of later chapters): 

Actually, as a divisibility rule it doesn't matter if you take the alternating sum "right to left" or "left to right". For example, with four digits it is okay to do $a-b+c-d$ or $d-c+b-a$. That is because the two alternating sums differ by $-1$, and thus,$$11 \mid (a-b+c-d) \text{ if and only if } 11 \mid d-c+b-a.$$So why do we want to stress doing the alternating sum "right to left"? What we'll show later is that actually this alternating sum also preserves the remainder when dividing by $11$.

Going back to our 4-digit example, this means that the remainder of $n=\overline{abcd}$ when dividing by $11$ is the same as the remainder when $d-c+b-a$ when dividing by $11$. Here, multiplying by $-1$ DOES change the remainder, so this order is important.

Hope this helps!