I don't think it's possible to derive these identities from quadratic transformations. However, it is possible to derive identities with $z=1/4$ from cubic transformations, though I don't know if the specific identity that the questioner asks about can be proved this way.
Here's an example. In the cubic transformation
\begin{multline*}{}_{2}F_{1}\left({{3a,1/3-a}\atop 2a+5/6}\Bigm |z\right)\\ =\left( 1-z \right) ^{-a} \left( 1+8z \right) ^{-2a} {}_{2}F_{ 1}\left({{a,a+1/2}\atop 2a+5/6}\Bigm |{\frac {27 z}{ \left(1-z \right) \left( 1+8z \right) ^{2}}}\right) \end{multline*}\begin{multline*}{}_{2}F_{1}\left({{3a,1/3-a}\atop 2a+5/6}\Bigm |z\right)\\
=\left( 1-z \right) ^{-a}
\left( 1+8z \right) ^{-2a}
{}_{2}F_{
1}\left({{a,a+1/2}\atop 2a+5/6}\Bigm |{\frac {27 z}{ \left(1-z
\right) \left( 1+8z \right) ^{2}}}\right) \end{multline*}
we can set $z=1/4$ to get
\begin{align*} {}_{2}F_{1}\left({{3a,1/3-a}\atop 2a+5/6}\Bigm |1/4\right)&= \left( {\frac {4}{27}} \right) ^{a} {}_{2} F_{1}\left({{a,a+1/2}\atop 2a+5/6}\Bigm |1\right)\\ &= \frac{2\pi}{\sqrt3} \left( {\frac {4}{27}} \right) ^{a}{\frac {\Gamma \left( 2a+5/6 \right) }{\Gamma \left( 2/3 \right) \Gamma \left( a+5/6 \right) \Gamma \left( a+1/3 \right) }}. \end{align*}\begin{align*}
{}_{2}F_{1}\left({{3a,1/3-a}\atop 2a+5/6}\Bigm |1/4\right)&=
\left( {\frac {4}{27}} \right) ^{a}
{}_{2}
F_{1}\left({{a,a+1/2}\atop 2a+5/6}\Bigm |1\right)\\
&=
\frac{2\pi}{\sqrt3} \left( {\frac {4}{27}} \right) ^{a}{\frac {\Gamma \left( 2a+5/6 \right) }{\Gamma \left( 2/3 \right)
\Gamma \left( a+5/6 \right) \Gamma \left( a+1/3 \right) }}.
\end{align*}
