The sum to find can be written in the two equivalent forms, as demonstrated
in the precedent post.
$$
\eqalign{
& S(L,q) = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{\Gamma (L + r - 2q)}
\over {\Gamma (L + r - 1 + 2q)}}{{\Gamma (L + r + l - 1 + 2q)} \over {\Gamma (L + r + l + 2)}}{{r + 1} \over {r + l + 2}}} } = \cr
& = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {\left( {r + 1} \right)\left( {L + r - 1 + 2q} \right)^{\,\overline {\, - \left( { 4q-1} \right)\,} }
\left( {L + r + l + 2} \right)^{\,\overline {\, - 3 + 2q\,} } {1 \over {r + l + 2}}} } \cr
& = \sum\limits_{s = 0}^\infty {{1 \over {s + 2}}\left( {L + s + 2} \right)^{\,\overline {\, - 3 + 2q\,} }
\sum\limits_{r = 0}^s {\left( {r + 1} \right)\left( {L + r - 1 + 2q} \right)^{\,\overline {\, - \left( { 4q-1} \right)\,} } } } \cr}
$$
Now, using the composition formula for the Rising Factorial
$$
x^{\,\overline {\,a + b\,} } = x^{\,\overline {\,a\,} } \left( {x + a} \right)^{\,\overline {\,b\,} }
$$
we obtain two fundamental blocks that we need for proceeding
$$
\eqalign{
& \left( {r + 1} \right)\left( {r + 1 + a} \right)^{\,\overline {\,1 - b\,} }
= {{\left( {r + 1} \right)^{\,\overline {\,1\,} } \left( {r + 2} \right)^{\,\overline {\,a - 1\,} } \left( {r + 1 + a} \right)^{\,\overline {\,1 - b\,} } }
\over {\left( {r + 2} \right)^{\,\overline {\,a - 1\,} } }} = \cr
& = {{\left( {r + 1} \right)^{\,\overline {\,a + 1 - b\,} } } \over {\left( {r + 2} \right)^{\,\overline {\,a - 1\,} } }}
= {{1^{\,\overline {\,r + 1\,} } } \over {1^{\,\overline {\,r\,} } }}{{1^{\,\overline {\,r\,} } \left( {r + 1} \right)^{\,\overline {\,a + 1 - b\,} } }
\over {1^{\,\overline {\,r + 1\,} } \left( {r + 2} \right)^{\,\overline {\,a - 1\,} } }}
= {{2^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } }}{{1^{\,\overline {\,r + a + 1 - b\,} } } \over {1^{\,\overline {\,r + a\,} } }} = \cr
& = {{2^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } }}{{1^{\,\overline {\,a + 1 - b\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } }
\over {1^{\,\overline {\,a\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}
= \left( {a + 1} \right)^{\,\overline {\,1 - b\,} } {{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } }
\over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }} \cr}
$$
and
$$
\eqalign{
& {1 \over {y + 2}}\left( {y + 2 + c} \right)^{\,\overline {\,d\,} }
= {{\left( {y + 3} \right)^{\,\overline {\,c - 1\,} } \left( {y + 2 + c} \right)^{\,\overline {\, - d\,} } }
\over {\left( {y + 2} \right)^{\,\overline {\,1\,} } \left( {y + 3} \right)^{\,\overline {\,c - 1\,} } }} = \cr
& = {{\left( {y + 3} \right)^{\,\overline {\,c - d - 1\,} } } \over {\left( {y + 2} \right)^{\,\overline {\,c\,} } }}
= {{1^{\,\overline {\,y + 1\,} } } \over {1^{\,\overline {\,y + 2\,} } }}{{1^{\,\overline {\,y + 2\,} } \left( {y + 3} \right)^{\,\overline {\,c - d - 1\,} } }
\over {1^{\,\overline {\,y + 1\,} } \left( {y + 2} \right)^{\,\overline {\,c\,} } }} = \cr
& = {{1^{\,\overline {\,y + 1\,} } } \over {1^{\,\overline {\,y + 2\,} } }}{{1^{\,\overline {\,y + 1 + c - d\,} } } \over {1^{\,\overline {\,y + 1 + c\,} } }}
= {{1^{\,\overline {\,1\,} } 2^{\,\overline {\,y\,} } } \over {1^{\,\overline {\,2\,} } 3^{\,\overline {\,y\,} } }}
{{1^{\,\overline {\,1 + c - d\,} } \left( {2 + c - d} \right)^{\,\overline {\,y\,} } }
\over {1^{\,\overline {\,1 + c\,} } \left( {2 + c} \right)^{\,\overline {\,y\,} } }} = \cr
& = {{1^{\,\overline {\,1\,} } 1^{\,\overline {\,1 + c - d\,} } } \over {1^{\,\overline {\,2\,} } 1^{\,\overline {\,1 + c\,} } }}
{{2^{\,\overline {\,y\,} } \left( {2 + c - d} \right)^{\,\overline {\,y\,} } } \over {3^{\,\overline {\,y\,} } \left( {2 + c} \right)^{\,\overline {\,y\,} } }} = \cr
& = {{\left( {c + 2} \right)^{\,\overline {\, - d\,} } } \over 2}{{2^{\,\overline {\,y\,} } \left( {2 + c - d} \right)^{\,\overline {\,y\,} } }
\over {3^{\,\overline {\,y\,} } \left( {2 + c} \right)^{\,\overline {\,y\,} } }} \cr}
$$
1) the $(r,l)$ version
Applying the above to the sum in $(r,l)$
$$ \bbox[lightyellow] {
\eqalign{
& \left\{ \matrix{
\matrix{ {a = L - 2 + 2q} \hfill & {b = \,4q} \hfill \cr {c = L} \hfill & {d = 3 - 2q\,} \hfill \cr } \hfill \cr
A = {1 \over 2}\left( {c + 2} \right)^{\,\overline {\, - d\,} } \left( {a + 1} \right)^{\,\overline {\,1 - b\,} }
= {1 \over 2}\left( {L + 2} \right)^{\,\overline {\, - 2 - 2q\,} } \hfill \cr} \right. \cr
& S(L,q) = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty
{\left( {r + 1} \right)\left( {r + 1 + L - 2 + 2q} \right)^{\,\overline {\,1 - 4q\,} } \left( {L + r + l + 2} \right)^{\,\overline {\, - \left( {3 - 2q} \right)\,} }
{1 \over {r + l + 2}}} } = \cr
& = A\;\sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{2^{\,\overline {\,r + l\,} } \left( {2 + c - d} \right)^{\,\overline {\,r + l\,} } }
\over {3^{\,\overline {\,r + l\,} } \left( {2 + c} \right)^{\,\overline {\,r + l\,} } }}{{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } }
\over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} } \cr}
} \tag{1.1}$$
Now, the above, once rewritten as
$$ \bbox[lightyellow] {
\eqalign{
& S(L,q) = A\;\sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{2^{\,\overline {\,r + l\,} } \left( {2 + c - d} \right)^{\,\overline {\,r + l\,} } }
\over {3^{\,\overline {\,r + l\,} } \left( {2 + c} \right)^{\,\overline {\,r + l\,} } }}
{{2^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } }
\over {\left( {a + 1} \right)^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } }}} } {{x^{\,r} } \over {r!}}{{y^{\,l} } \over {l!}} = \cr
& = A\,F\left( {\matrix{ 2 \cr 3 \cr 2 \cr 2 \cr
} \,\left| {\,\matrix{
{2;\left( {2 + c - d} \right)} \cr
{2,\,1\;;\;\left( {a + 2 - b} \right),1\;} \cr
{3;\left( {2 + c} \right)} \cr
{\;\left( {a + 1} \right),1} \cr
} \,} \right|x,y} \right)\quad \quad \left| {\;x = y = 1} \right. \cr}
} \tag{1.2}$$
tells us that the sum is a
Kampé de Fériet function computed at $x=y=1$.
Looking on the web one can find various specialized papers dealing with the
expansion of Kampé de Fériet function function in terms of Hypergeometric functions.
So the expansion found by @Nikos Bagis well enters into the scene.
(re. to e.g. this paper and to this and this).
Also interesting is the connection with the "Extended Beta" integral representation
(re. for instance to this and this paper, and specially this), which relates with the integral representation
hinted in my answer to the previous post
2) the $(r,s)$ truncated sum version
Restarting from (1.1) above, and writing $S(L,q)$ in the truncated way we get
$$ \bbox[lightyellow] {
\eqalign{
& S(L,q) = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty
{\left( {r + 1} \right)\left( {r + 1 + L - 2 + 2q} \right)^{\,\overline {\,1 - 4q\,} } \left( {L + r + l + 2} \right)^{\,\overline {\, - \left( {3 - 2q} \right)\,} }
{1 \over {r + l + 2}}} } = \cr
& = A\;\sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{2^{\,\overline {\,r + l\,} } \left( {2 + c - d} \right)^{\,\overline {\,r + l\,} } }
\over {3^{\,\overline {\,r + l\,} } \left( {2 + c} \right)^{\,\overline {\,r + l\,} } }}{{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } }
\over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} } = \cr
& = A\;\sum\limits_{s = r + l = 0}^\infty {{{2^{\,\overline {\,s\,} } \left( {2 + c - d} \right)^{\,\overline {\,s\,} } }
\over {3^{\,\overline {\,s\,} } \left( {2 + c} \right)^{\,\overline {\,s\,} } }}\sum\limits_{r = 0}^s {{{2^{\,\overline {\,r\,} }
\left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} } \cr}
} \tag{2.1}$$
In order to reduce it, exactly or asymptotically, into more manageable terms,
I am currently exploring the possibilities offered by some possible approaches that I could individuate up to now.
2.a) Product of two Hypergeometric
A truncated Hypergeometric series can be transformed into the limit of a standard one, by adding
two additional parameters
$$
\eqalign{
& \sum\limits_{r = 0}^s {{{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } }
\over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \sum\limits_{r = 0}^\infty
{{{\left( { - s} \right)^{\,\overline {\,r\,} } 2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } }
\over {\left( { - s + \varepsilon } \right)^{\,\overline {\,r\,} } 1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} = \cr
& = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \sum\limits_{r = 0}^\infty {{{\left( { - s} \right)^{\,\overline {\,\varepsilon \,} }
2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {\left( { - s + r} \right)^{\,\overline {\,\varepsilon \,} }
1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} \cr}
$$
(re. to the renowned "Concrete Mathematics" - Ch. 5 - ex. 34)
which takes advantage of that $\left( { - s} \right)^{\,\overline {\,r\,} }$ becomes null for non negative integers $s < r$.
This gives us a summand product of two hypergeometric terms, but with one fraction containing both parameters.
Since
$$
{{\left( { - s} \right)^{\,\overline {\,\varepsilon \,} } } \over {\left( { - s + r} \right)^{\,\overline {\,\varepsilon \,} } }}
= {{\Gamma \left( { - s + \varepsilon } \right)\Gamma \left( { - s + r} \right)} \over {\Gamma \left( { - s} \right)\Gamma
\left( { - s + r + \varepsilon } \right)}}
= {{{\rm B}\left( { - s + r,\varepsilon } \right)} \over {{\rm B}\left( { - s,\varepsilon } \right)}}
$$
there might be a possibility to expand it in some interesting way.
2.b) series expression of the truncated Hypergeometric
Trying to apply the formulas provided in this work
