For your first question, the answer is yes, you can change indeed rewrite your outer sum by substituting $m-1$ for $j$ and changing the limits of that sum.
You can't, however, make the step indicated in your second question. Consider, for example, your middle summation. That sum is over all pairs $(m, j)$ satisfying by the predicate
$$
P(m, j) = (1\le m\le 8) \land (1\le j\le 8) \land (j\ne m-1)
$$
There will $64-8 = 56$ terms in this sum. The first step in your argument was correct:
$$
\sum_{P(m, j)} 8 = 8\left(\sum_{P(m, j)} 1\right)
$$
but your next step wasn't, since
$$
\sum_{P(m,j)} 1 \ne \sum_1^8 1
$$
To see this, just observe that the sum on the left has 56 terms and the sum on the right has only 8.
Finally, a useful technique for problems like this is to ignore for the moment the excluded diagonal terms, sum all the terms, and then subtract the excluded ones. Using your chessboard idea, you noticed that the squares in the chessboard contain the numbers $1, \dots , 64$ so the sum of all the entries is
$$
\frac{64\cdot65}{2} = 2080
$$
Now all we have to do is subtract the sum of the 7 diagonal elements in positions
$$
(k-1, k) = (2, 1), (3, 2), (4, 3), (5, 4), (6, 5), (7, 6), (8, 7)
$$
where the values are $9, 18, 27, 36, 45, 54, 63$. The sum of these is 252, so the answer to the problem is $2080-252=1828$.
You could calculate as follows
\begin{align*}
2\sum_{n=0}^\infty& a_nx^{n+1} +\sum_{n=1}^ \infty nb_nx^{n-1}\\
&=2\sum_{n=1}^\infty a_{n-1}x^{n} +\sum_{n=0}^ \infty (n+1)b_{n+1}x^{n}\tag{1}\\
&=2\sum_{n=1}^\infty a_{n-1}x^{n} +\left(b_1+\sum_{n=1}^ \infty (n+1)b_{n+1}x^{n}\right)\tag{2}\\
&=b_1+\sum_{n=1}^{\infty}\left(2a_{n-1}+(n+1)b_{n+1}\right)x^n\tag{3}
\end{align*}
Comment:
In (1) we shift the index of both series by one in order to obtain summands with $x^n$.
In (2) we separate the first summand of the second series. So we get the same index starting value $n=1$ in both series.
In (3) we collect the terms into one series.
If we denote the series (3) with
\begin{align*}
C(x)=\sum_{n=0}^\infty c_nx^n
\end{align*}
we observe
\begin{align*}
c_0&=b_1\\
c_n&=2a_{n-1}+(n+1)b_{n+1}\qquad\qquad n\geq 1
\end{align*}
Best Answer
All he's doing is reversing the order of summation. Try writing it out. Originally, $i$ ranges from $1$ to $n$, so our summation is: $$ S_n = 1 + 2 + \cdots + i + \cdots + n $$ After making the substitution $i = n - j \iff j = n - i$, we note that $j$ now ranges backwards from $n-1$ to $n-n$, so our summation is: $$ S_n = [n-(n-1)] + [n-(n-2)] + \cdots + [n-(n-j)] + \cdots + [n - (n-n)] $$ By simplifying the terms in the round brackets and reversing the order of summation, we obtain: $$ S_n = [n - 0] + [n-1] + \cdots + [n-j] + \cdots + [n - (n - 1)] $$ so that now $j$ is ranging from $0$ to $n - 1$.