Remark. What follows answers one of several queries that appeared
at this post, which each query replacing the previous one. We suggest
making a list so that all the different varieties may be examined.
Starting from the claim
$$\bbox[5px,border:2px solid #00A000]{
{m+n\choose s+1} - {n\choose s+1}
= \sum_{q=0}^s \frac{m}{q+1}
{m+1+2q\choose q} {n-2-2q\choose s-q}}$$
we observe that
$${m+1+2q\choose q+1} - {m+1+2q\choose q}
\\ = \frac{m+1+q}{q+1} {m+1+2q\choose q} - {m+1+2q\choose q}
\\ = \frac{m}{q+1} {m+1+2q\choose q}.$$
Therefore we have two sums,
$$\sum_{q=0}^s
{m+1+2q\choose q+1} {n-2-2q\choose s-q}
- \sum_{q=0}^s
{m+1+2q\choose q} {n-2-2q\choose s-q}.$$
For the first one we write
$$\sum_{q=0}^s
[w^{q+1}] (1+w)^{m+1+2q} [z^{s-q}] (1+z)^{n-2-2q}
\\ = \mathrm{Res}_{w=0}
(1+w)^{m+1} [z^s] (1+z)^{n-2}
\sum_{q=0}^s \frac{1}{w^{q+2}} z^q (1+w)^{2q} (1+z)^{-2q}.$$
We may extend $q$ beyond $s$ because of the coefficient extractor
$[z^s]$ in front, getting
$$ \mathrm{Res}_{w=0} \frac{1}{w^2} (1+w)^{m+1} [z^s] (1+z)^{n-2}
\sum_{q\ge 0} z^q w^{-q} (1+w)^{2q} (1+z)^{-2q}
\\ = \mathrm{Res}_{w=0} (1+w)^{m+1} [z^s] (1+z)^{n-2}
\frac{1}{w^2} \frac{1}{1-z(1+w)^2/w/(1+z)^2}
\\ = \mathrm{Res}_{w=0} (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{w} \frac{1}{w(1+z)^2-z(1+w)^2}.$$
Repeat the calculation for the second one to get
$$\mathrm{Res}_{w=0} (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{w(1+z)^2-z(1+w)^2}.$$
Now we have
$$\left(\frac{1}{w}-1\right)\frac{1}{w(1+z)^2-z(1+w)^2}
= \frac{1}{w-z} \frac{1}{w(1+w)}
- \frac{1}{1-wz} \frac{1}{1+w}
\\ = \frac{1}{1-z/w} \frac{1}{w^2(1+w)}
- \frac{1}{1-wz} \frac{1}{1+w}.$$
We thus obtain two components, the first is
$$\mathrm{Res}_{w=0} (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{1-z/w} \frac{1}{w^2(1+w)}
\\ = \mathrm{Res}_{w=0} \frac{1}{w^2} (1+w)^{m}
[z^s] (1+z)^{n} \frac{1}{1-z/w}
\\ = \mathrm{Res}_{w=0} \frac{1}{w^2} (1+w)^{m}
\sum_{q=0}^s {n\choose q} \frac{1}{w^{s-q}}
= \sum_{q=0}^s {n\choose q}
\mathrm{Res}_{w=0} \frac{1}{w^{s-q+2}} (1+w)^{m}
\\ = \sum_{q=0}^s {n\choose q} [w^{s-q+1}] (1+w)^m
= [w^{s+1}] (1+w)^m \sum_{q=0}^s {n\choose q} w^q
\\ = - {n\choose s+1}
+ [w^{s+1}] (1+w)^m \sum_{q=0}^{s+1} {n\choose q} w^q.$$
We may extend $q$ beyond $s+1$ due to the coefficient extractor in
front, to get
$$- {n\choose s+1}
+ [w^{s+1}] (1+w)^m \sum_{q\ge 0} {n\choose q} w^q
= - {n\choose s+1}
+ [w^{s+1}] (1+w)^{m+n}$$
This is
$$\bbox[5px,border:2px solid #00A000]{
{m+n\choose s+1} - {n\choose s+1}.}$$
We have the claim, so we just need to prove that the second component
will produce zero. We obtain
$$\mathrm{Res}_{w=0} (1+w)^{m+1} [z^s] (1+z)^{n}
\frac{1}{1-wz} \frac{1}{1+w}
\\ = \mathrm{Res}_{w=0} (1+w)^{m} [z^s] (1+z)^{n}
\frac{1}{1-wz}
\\ = \mathrm{Res}_{w=0} (1+w)^{m}
\sum_{q=0}^s {n\choose q} w^{s-q}
= \sum_{q=0}^s {n\choose q} \mathrm{Res}_{w=0} w^{s-q} (1+w)^{m}
= 0.$$
This concludes the argument. Having reached the end of the computation
we observe that we did not require the full mechanics of the complex
residue and a coefficient extractor would have sufficed.