Note that we can rewrite the problem as
$$
\eqalign{
& \left\{ \matrix{
n_{\,1} < n_{\,2} < n_{\,3} < \cdots < n_{\,q - 1} < n_{\,q} \hfill \cr
n_{\,1} + n_{\,2} + n_{\,3} + \cdots + n_{\,q - 1} + n_{\,q} = s \hfill \cr} \right. \cr
& \quad \quad \Downarrow \cr
& \left\{ \matrix{
n_{\,1} \le n_{\,2} - 1 \le n_{\,3} - 2 \le \cdots \le n_{\,q - 1} - \left( {q - 2} \right) \le n_{\,q} - \left( {q - 1} \right) \hfill \cr
n_{\,1} + \left( {n_{\,2} - 1} \right) + \cdots + \left( {n_{\,q} - \left( {q - 1} \right)} \right) = s - \left( \matrix{
q \cr
2 \cr} \right) \hfill \cr} \right. \cr
& \quad \quad \Downarrow \cr
& \left\{ \matrix{
m_{\,1} \le m_{\,2} \le \cdots \le m_{\,q - 1} \le m_{\,q} \hfill \cr
m_{\,1} + m_{\,2} + \cdots + m_{\,q - 1} + m_{\,q} = s - \left( \matrix{
q \cr
2 \cr} \right) \hfill \cr} \right. \cr}
$$
That means that:
- if $1 \le m_1$, which means $ m_k \in \mathbb N$, then the number you are looking for is the
number of partitions of $s -\binom{q}{2}$ into $q$ parts;
- if $0 \le m_1$, which means $ 0 \le m_k \in \mathbb Z$, then the number you are looking for is the
number of partitions of $s -\binom{q}{2}$ into at most $q$ parts.
You can then refer to this Wikipedia article about partitions with restricted part size/number and to the vast literature on the subject.
Keeping instead with your approach, which is a valid alternative, we have
$$
\eqalign{
& \left\{ \matrix{
0 < n_{\,1} < n_{\,2} < n_{\,3} < \cdots < n_{\,q - 1} < n_{\,q} \hfill \cr
n_{\,1} + n_{\,2} + n_{\,3} + \cdots + n_{\,q - 1} + n_{\,q} = s \hfill \cr} \right. \cr
& \quad \quad \Downarrow \cr
& \left\{ \matrix{
1 \le n_{\,1} = m_{\,1} \hfill \cr
1 \le m_{\,k} = n_{\,k} - n_{\,k - 1} \quad \left| {\;2 \le k \le q} \right. \hfill \cr
qm_{\,1} + \left( {q - 1} \right)m_{\,2} + \cdots + 2m_{\,q - 1} + 1m_{\,q} = s \hfill \cr} \right. \cr
& \quad \quad \Downarrow \cr
& \left\{ \matrix{
0 \le p_{\,q + 1 - k} = m_{\,k} - 1 \hfill \cr
1p_{\,1} + 2p_{\,2} + \cdots + q\,p_{\,q} = s - \left( \matrix{
q + 1 \cr
2 \cr} \right) \hfill \cr} \right. \cr}
$$
In your example with $q=5$ we have that, if we take the polynomial
$$
\eqalign{
& P(x) = \left( {x^{\,1} \cdot x^{\,2} \cdot \cdots \cdot x^{\,5} } \right)\underbrace {\left( {x^{\,1} + x^{\,2} + \cdots + x^{\,5} } \right)\left( {x^{\,1} + x^{\,2} + \cdots + x^{\,5} } \right) \cdots \left( {x^{\,1} + x^{\,2} + \cdots + x^{\,5} } \right)}_{s - 1\, \le \,t\,{\rm terms}} \cr
& = \cdots + x^{\left( {\scriptstyle 6 \atop
\scriptstyle 2} \right)} x^{\,1\,k_{\,1} } x^{\,2\,k_{\,2} } \cdots x^{\,5\,k_{\,5} } + \cdots \quad \left| {\;0 \le k_{\,1} + k_{\,2} + \cdots + k_{\,5} = t} \right. \cr}
$$
we do get
$$
\left[ {x^{\,s} } \right]P(x) = {\rm number}\,{\rm of}\,{\rm solutions}\left\{ \matrix{
1 \le \left( {k_{\,j} + 1} \right) \hfill \cr
1\,\left( {k_{\,1} + 1} \right) + 2\,\left( {k_{\,2} + 1} \right) + \cdots + 5\left( {k_{\,5} + 1} \right) = s \hfill \cr} \right.
$$
Instead of the above, especially for analysis purposes, we have better to consider the fractional function (which has an infinite power expansion)
$$
F(x) = {x \over {1 - x}}{{x^{\,2} } \over {1 - x^{\,2} }} \cdots {{x^{\,5} } \over {1 - x^{\,5} }} = x^{\left( {\scriptstyle 6 \atop
\scriptstyle 2} \right)} {1 \over {1 - x}}{1 \over {1 - x^{\,2} }} \cdots {1 \over {1 - x^{\,5} }}
$$
and this seems to be what your book is suggesting.