Another approach comes from expressing the sums in $(s,r)$ and then managing the resulting partial sum.
1) the sum $F_1$
Starting from your first question, we write it as
$$ \bbox[lightyellow] {
\eqalign{
& F_{\,1} (L,q,x) = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty
{{{\Gamma (L + r - 2q)} \over {\Gamma (L + r - 1 + 2q)}}{{\Gamma (L + r + l - 1 + 2q)} \over {\Gamma (L + r + l + 2)}}x^{\,l} x^{\,r} } } = \cr
& = \sum\limits_{s = 0}^\infty {{{\Gamma (L + s - 1 + 2q)}
\over {\Gamma (L + s + 2)}}x^{\,s} } \sum\limits_{r = 0}^s {{{\Gamma (L + r - 2q)} \over {\Gamma (L + r - 1 + 2q)}}} = \cr
& = \sum\limits_{s = 0}^\infty {{{\Gamma (L - 1 + 2q)} \over {\Gamma (L + 2)}}{{\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }
\over {\left( {L + 2} \right)^{\,\overline {\,s\,} } }}x^{\,s} } {{\Gamma \left( {L - 2q} \right)} \over {\Gamma \left( {L - 1 + 2q} \right)}}
\sum\limits_{r = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,r\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,r\,} } }}} = \cr
& = {{\Gamma \left( {L - 2q} \right)} \over {\Gamma (L + 2)}}\sum\limits_{s = 0}^\infty {{{\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }
\over {\left( {L + 2} \right)^{\,\overline {\,s\,} } }}x^{\,s} } \sum\limits_{r = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,r\,} } }
\over {\left( {L - 1 + 2q} \right)^{\,\overline {\,r\,} } }}} \cr}
} \tag {1.0}$$
Let us consider the inner partial sum.
It is a sum of hypergeometric terms and to it we can apply the Gosper's algorithm,
which I find to be well described and analyzed in the renowned book Concrete Mathematics.
In the following I am concisely developing the steps recommended in
the relevant "Ch. 5.7 - Partial Hypergeometric Sums": please refer to that for further details.
So
- we write the sum as
$$
S_P (L,q,s) = \sum\limits_{k = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }}}
= \sum\limits_{k = 0}^s {t_{\,k} }
$$
- express the ratio of the summand as
$$
{{t_{\,k + 1} } \over {t_{\,k} }} = {{\left( {L - 2q + k} \right)} \over {\left( {L - 2 + 2q + k + 1} \right)}} = {{p(k + 1)} \over {p(k)}}{{q(k)} \over {r(k + 1)}}
$$
- choose the three polynomials to be
$$
\left\{ \matrix{ p(k) = 1 \hfill \cr q(k) = L - 2q + k \hfill \cr r(k) = L - 2 + 2q + k \hfill \cr} \right.
$$
- determine the polynomial $s(k)$ by solving
$$
p(k) = q(k)s(k + 1) - r(k)s(k)\quad \Rightarrow \quad s(k) = {1 \over {4q - 2}}
$$
- arrive to find the antidifference $T(k)$ of $t_k$ to be
$$ \bbox[lightyellow] {
\left\{ \matrix{
T(k + 1) - T(k) = t_{\,k} \hfill \cr
T(k) = {{r(k)s(k)t(k)} \over {p(k)}} = {{L - 2 + 2q} \over {4q - 2}}{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } }
\over {\left( {L - 2 + 2q} \right)^{\,\overline {\,k\,} } }} \hfill \cr} \right.
} \tag {1.1}$$
- and conclude that
$$ \bbox[lightyellow] {
\eqalign{
& S_P (L,q,s) = \sum\limits_{k = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }}}
= T(s + 1) - T(0) = \cr
& = {{L - 2 + 2q} \over {2 - 4q}}\left( {{{\left( {L - 2q} \right)^{\,\overline {\,s + 1\,} } }
\over {\left( {L - 2 + 2q} \right)^{\,\overline {\,s + 1\,} } }} - 1} \right) = \cr
& = {{L - 2 + 2q} \over {2 - 4q}}\left( {{{\left( {L - 2q} \right)\left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } }
\over {\left( {L - 2 + 2q} \right)\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} - 1} \right) = \cr
& = \left( {{{L - 2q} \over {2 - 4q}}} \right){{\left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }}
- {{L - 2 + 2q} \over {2 - 4q}} \cr}
} \tag {1.2}$$
Note that, although $T(k)$ has a pole at $q=1/2$, it cancels out in the difference $T(k+1)-T(k)=t_k$.
Upon inserting (1.2) back into (1.0), then it is just a matter of algebraic manipulation.
2) the sum $F_2$
Since $F_2$ is same as $F_1$ with the inner sum being multiplicated by $r$, then this is rewritten as
$$ \bbox[lightyellow] {
\eqalign{
& S_P (L,q,s) = \sum\limits_{k = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }}k} = \cr
& = \sum\limits_{k = 1}^s {{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }}k}
= \sum\limits_{k = 1}^s {{{\left( {L-2q} \right)^{\,\overline {\,k + 1\,} } } \over {\left( {L-1+2q} \right)^{\,\overline {\,k+1\,} } }}\left( {k+1} \right)} = \cr
& = {{\left( {L - 2q} \right)} \over {\left( {L - 1 + 2q} \right)}}\sum\limits_{k = 0}^{s - 1}
{{{\left( {L - 2q + 1} \right)^{\,\overline {\,k\,} } } \over {\left( {L + 2q} \right)^{\,\overline {\,k\,} } }}\left( {k + 1} \right)} \cr}
} \tag {2.0}$$
Proceeding similarly to the above, the steps in summary being
$$
\eqalign{
& \left\{ \matrix{
p(k) = \left( {k + 1} \right) \hfill \cr
q(k) = \left( {L - \left( {2q - 1} \right) + k} \right) \hfill \cr
r(k) = \left( {L + \left( {2q - 1} \right) + k} \right) \hfill \cr} \right. \cr
& s(k) = {1 \over {3 - 4q}}\left( {k + {{L - 2 + 2q} \over {2\left( {2q - 1} \right)}}} \right) \cr}
$$
we reach to
$$ \bbox[lightyellow] {
\eqalign{
& T(k) = {{r(k)s(k)t(k)} \over {p(k)}} = \cr
& = {{L - 1 + 2q} \over {3 - 4q}}\left( {k + {{L - 2 + 2q} \over {4q - 2}}} \right){{\left( {L + 1 - 2q} \right)^{\,\overline {\,k\,} } }
\over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }} \cr}
} \tag {2.1}$$
Thus
$$ \bbox[lightyellow] {
\eqalign{
& S_P (L,q,s) = \sum\limits_{k = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }}k} = \cr
& = {{\left( {L - 2q} \right)} \over {\left( {L - 1 + 2q} \right)}}\sum\limits_{k = 0}^{s - 1} {{{\left( {L - 2q + 1} \right)^{\,\overline {\,k\,} } }
\over {\left( {L + 2q} \right)^{\,\overline {\,k\,} } }}\left( {k + 1} \right)} = \cr
& = {{\left( {L - 2q} \right)} \over {3 - 4q}}\left( {\left( {s + {{L - 2 + 2q} \over {4q - 2}}} \right){{\left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } }
\over {\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} - \left( {{{L - 2 + 2q} \over {4q - 2}}} \right)} \right) = \cr
& = {{\left( {L - 2q} \right)} \over {3 - 4q}}\left( {\left( {s + 1 + {{L - 2q} \over {4q - 2}}} \right){{\left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } }
\over {\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} - \left( {{{L - 2 + 2q} \over {4q - 2}}} \right)} \right) = \cr
& = {{\left( {L - 2q} \right)} \over {3 - 4q}}\left( {{{2^{\,\overline {\,s\,} } \left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } } \over {1^{\,\overline {\,s\,} }
\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} + \left( {{{L - 2q} \over {4q - 2}}} \right){{\left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } }
\over {\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} - \left( {{{L - 2 + 2q} \over {4q - 2}}} \right)} \right) \cr}
} \tag {2.2}$$
3) Gosper's algorithm applied directly to the original sum
The Gosper's algorithm can be applied as well
to your original sum given in the precedent post, in reply to which
it was already expressed in closed form as a
Kampé de Fériet function computed at $x=y=1$.
Following the same steps as above, we can now express it also as a linear combination of hypergeometric functions.
In fact, putting
$$ \bbox[lightyellow] {
\eqalign{
& \left\{ \matrix{
\matrix{
{a = L - 2 + 2q} \hfill & {b = \,4q} \hfill \cr
{c = L} \hfill & {d = 3 - 2q\,} \hfill \cr
} \hfill \cr
A = {1 \over 2}\left( {c + 2} \right)^{\,\overline {\, - d\,} } \left( {a + 1} \right)^{\,\overline {\,1 - b\,} }
= {1 \over 2}\left( {L + 2} \right)^{\,\overline {\, - 2 - 2q\,} } \hfill \cr} \right. \cr
& S(L,q) = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{\Gamma (L + r - 2q)} \over {\Gamma (L + r - 1 + 2q)}}
{{\Gamma (L + r + l - 1 + 2q)} \over {\Gamma (L + r + l + 2)}}{{r + 1} \over {r + l + 2}}} } \cr
& = A\;\sum\limits_{s = 0}^\infty {{{2^{\,\overline {\,s\,} } \left( {2 + c - d} \right)^{\,\overline {\,s\,} } }
\over {3^{\,\overline {\,s\,} } \left( {2 + c} \right)^{\,\overline {\,s\,} } }}
\sum\limits_{r = 0}^s {{{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } }
\over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} } \cr}
} \tag {3.0}$$
for the inner sum
$$
S_{\,p} (a,b,s) = \sum\limits_{k = 0}^s {{{2^{\,\overline {\,k\,} } \left( {a + 2 - b} \right)^{\,\overline {\,k\,} } } \over {1^{\,\overline {\,k\,} } \left( {a + 1} \right)^{\,\overline {\,k\,} } }}}
$$
we go through steps similar to the above and reach to
$$ \bbox[lightyellow] {
\eqalign{
& S_{\,p} (a,b,s) = \sum\limits_{k = 0}^s
{{{2^{\,\overline {\,k\,} } \left( {a + 2 - b} \right)^{\,\overline {\,k\,} } } \over {1^{\,\overline {\,k\,} } \left( {a + 1} \right)^{\,\overline {\,k\,} } }}} = \cr
& = {a \over {3 - b}}\left( {\left( {s + 1 - {{a - 1} \over {2 - b}}} \right){{\left( {a + 2 - b} \right)^{\,\overline {\,s + 1\,} } }
\over {a^{\,\overline {\,s + 1\,} } }} + {{a - 1} \over {2 - b}}} \right) = \cr
& = {a \over {3 - b}}\left( {{{\left( {a + 2 - b} \right)2^{\,\overline {\,s\,} } \left( {a + 3 - b} \right)^{\,\overline {\,s\,} } }
\over {a\;1^{\,\overline {\,s\,} } \left( {a + 1} \right)^{\,\overline {\,s\,} } }} - \left( {{{a - 1} \over {2 - b}}} \right){{\left( {a + 2 - b} \right)
\left( {a + 3 - b} \right)^{\,\overline {\,s\,} } } \over {a\left( {a + 1} \right)^{\,\overline {\,s\,} } }} + {{a - 1} \over {2 - b}}} \right) = \cr
& = {{\left( {a + 2 - b} \right)} \over {3 - b}}\left( {{{2^{\,\overline {\,s\,} } \left( {a + 3 - b} \right)^{\,\overline {\,s\,} } }
\over {1^{\,\overline {\,s\,} } \left( {a + 1} \right)^{\,\overline {\,s\,} } }} - \left( {{{a - 1} \over {2 - b}}} \right){{\left( {a + 3 - b} \right)^{\,\overline {\,s\,} } }
\over {\left( {a + 1} \right)^{\,\overline {\,s\,} } }} + {{a\left( {a - 1} \right)} \over {\left( {a + 2 - b} \right)\left( {2 - b} \right)}}} \right) \cr}
} \tag {3.1}$$
