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*} 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) }}.