I have worked out this proof up until the end, when the authors are combining terms here:
In the blue, they give an explanation for why you are able to do this simplification, but that wasn't sufficient enough for me. Would someone be able to expand on this?
On
This follows from the symmetry $$\binom{n}{m}=\binom{n}{n-m},$$ with $m\le n$ being nonnegative integers.
This is true because the number of selections of $m$ from a set of $n$ objects is exactly equal to the number of selecting the remaining $n-m$ instead, since each selection partitions this set into two, one part with $n,$ the other with $n-m.$ This gives the one-one correspondence between the selections.
In particular, for your case, we have $$\binom{m+1}{m+1}=\binom{m+1}{(m+1)-(m+1)}=\binom{m+1}{0},$$ which is equal to $1,$ of course. Thus the coefficients of first and last terms of the sum (both being $1$) can be replaced by binomial coefficients with the same form as those of the other (middle) terms, so that the whole sum can be compactly embraced under the single symbol $\sum.$
Recall $a^0=b^0=1$, so $$ a^{m+1}=\binom{m+1}{0} a^{m+1-0} b^0\quad\text{and}\quad b^{m+1}=\binom{m+1}{m+1} a^{m+1-(m+1)} b^{m+1}. $$ Hence you really have got all $k=0$ to $m+1$ covered.