We have that
$$
r_a(x) = {{\Gamma \left( {\left( {a + 1} \right)x} \right)} \over {\Gamma \left( x \right)}}
= {{\Gamma \left( {x + a\,x} \right)} \over {\Gamma \left( x \right)}} = x^{\,\overline {\,a\,x\,} }
$$
where $x^{\,\overline {\,y\,} } $ denotes the Rising Factorial.
Now, the Rising Factorial is defined (for $x$ and $y$ real and also complex) as
$$
h(x,y) = x^{\,\overline {\,y\,\,} } = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}} = \prod\nolimits_{\;k\, = \,\,0}^{\,y} {\left( {x + k} \right)}
$$
where the last term denotes the Indefinite Product, computed for $k$ ranging between the indicated bounds.
So we can write $f_a(x)$ as
$$ \bbox[lightyellow] {
f_{\,a} (x) = 2^{\,a\,x} {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}
= \prod\nolimits_{\;k\, = \,\,0}^{\,a\,x} {2\left( {x + k} \right)} = 2^{\,a\,x} x^{\,a\,x} \prod\nolimits_{\;k\, = \,\,0}^{\,a\,x} {\left( {1 + k/x} \right)}
} \tag{1}$$
Concerning the derivative of $r_a(x)$, since
$$
\left\{ \matrix{
{\partial \over {\partial x}}h(x,y) = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}}\left( {\psi \left( {x + y} \right)
- \psi \left( x \right)} \right) \hfill \cr
{\partial \over {\partial y}}h(x,y) = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}}\psi \left( {x + y} \right) \hfill \cr} \right.
$$
then, as you already found
$$
\eqalign{
& {d \over {dx}}r_a(x) = {\partial \over {\partial x}}h(x,y) + {\partial \over {\partial y}}h(x,y){d \over {dx}}y = \cr
& = {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\left( {\left( {a + 1} \right)\psi \left( {x + ax} \right)
- \psi \left( x \right)} \right) = \cr
& = {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\left( {a\psi \left( {x + ax} \right) + \left( {\psi \left( {x + ax} \right)
- \psi \left( x \right)} \right)} \right) \cr}
$$
where, for $0<x$ (and $0<a$) $\Gamma(x+ax)/\Gamma(x)$ is clearly positive.
However, while $\psi(x+ax)-\psi(x)$ is also positive since $\psi(x)$ is increasing in that range, $a\psi(a+ax)$ introduces a negative term for lower $x$.
To determine the limit of $r_a'(x)$ as $x \to 0^+$, let's consider the series development of
$$ \bbox[lightyellow] {
\left\{ \matrix{
\ln \Gamma (cx) = \ln \left( {{1 \over {cx}}} \right) - \gamma cx + O\left( {x^{\,2} } \right) \hfill \cr
\psi (cx) = - {1 \over {cx}} - \gamma + {{\pi ^{\,2} } \over 6}cx + O\left( {x^{\,2} } \right) = \hfill \cr
= - {1 \over {cx}} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + cx}}} \right)} \hfill \cr} \right.
} \tag{2}$$
Therefore
$$
\eqalign{
& \mathop {\lim }\limits_{x\; \to \;0^{\, + } } {d \over {dx}}r_a(x) = \mathop {\lim }\limits_{x\; \to \;0^{\, + } } {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\mathop {\lim }\limits_{x\; \to \;0^{\, + } } \left( {\left( {a + 1} \right)\psi \left( {x + ax} \right) - \psi \left( x \right)} \right) = \cr
& = {1 \over {a + 1}}\left( { - \gamma a} \right) \cr}
$$
which is negative for $0<a$.
Concerning $f_a(x)$ instead
$$
\eqalign{
& {d \over {dx}}f_{\,a} (x) = \;2^{\,a\,x} a\ln 2r_{\,a} (x) + 2^{\,a\,x} {d \over {dx}}r_{\,a} (x) = \cr
& = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + {d \over {dx}}\ln \left( {r_{\,a} (x)} \right)} \right) = \cr
& = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + {d \over {dx}}\ln \Gamma (x + ax) - {d \over {dx}}\ln \Gamma (x)} \right) = \cr
& = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + \left( {a + 1} \right)\psi (x + ax) - \psi (x)} \right) \cr}
$$
(which is the equation you already found)
and
$$ \bbox[lightyellow] {
\eqalign{
& \mathop {\lim }\limits_{x\; \to \;0^{\, + } } f_{\,a} '(x) = 1\left( {\left( {a\ln 2} \right){1 \over {a + 1}} - {{\gamma a} \over {a + 1}}} \right) = \cr
& = {a \over {a + 1}}\left( {\ln 2 - \gamma } \right) = {a \over {a + 1}}0.1159 \cdots \cr}
} \tag{3}$$
Proceeding with the development of the derivative above
$$ \bbox[lightyellow] {
\eqalign{
& {d \over {dx}}f_{\,a} (x)\;\mathop /\limits_{} \;\left( {2^{\,a\,x} r_{\,a} (x)} \right) = \cr
& = a\ln 2 + \left( {a + 1} \right)\psi (x + ax) - \psi (x) = \cr
& = a\ln 2 + \left( {a + 1} \right)\left( { - {1 \over {\left( {a + 1} \right)x}} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
- {1 \over {k + 1 + \left( {a + 1} \right)x}}} \right)} } \right) - \left( { - {1 \over x} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
- {1 \over {k + 1 + x}}} \right)} } \right) = \cr
& = \left( {a\ln 2 - {1 \over x} - \left( {a + 1} \right)\gamma + {1 \over x} + \gamma } \right)
+ \left( {a + 1} \right)\sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + \left( {a + 1} \right)x}}} \right)}
- \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + x}}} \right)} = \cr
& = a\left( {\ln 2 - \gamma } \right) + \sum\limits_{0\, \le \,k} {\left( {{a \over {k + 1}} + {1 \over {k + 1 + x}}
- {{\left( {a + 1} \right)} \over {k + 1 + \left( {a + 1} \right)x}}} \right)} = \cr
& = a\left( {\ln 2 - \gamma } \right) + a\sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
- {{k + 1} \over {\left( {k + 1 + x} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}}} \right)} \cr}
} \tag{4}$$
and
$$ \bbox[lightyellow] {
\eqalign{
& 0 \le {1 \over {k + 1}} - {{k + 1} \over {\left( {k + 1 + x} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}} = \cr
& = {1 \over {k + 1}} - {1 \over {\left( {1 + x/\left( {k + 1} \right)} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}}\quad \left| {\;0 \le x,k} \right. \cr}
} \tag{5}$$
therefore your thesis is demonstrated.
Best Answer
Firstly, $$\frac12\Gamma\left(\frac12\right)=\dfrac{\sqrt\pi}2\approx 0.886.\tag1$$
Denote $$y=\frac12\Gamma\left(\frac12\Gamma\left(\frac12\right)\right),\tag2$$ then $$y = \frac12\Gamma\left(\dfrac{1-x}{1+x}\right) < \frac12 + \gamma x + \left(\frac{\pi^2}6+\gamma^2-\gamma\right)x^2 + 0.251 x^3 + 2.33 x^4\tag3$$ (see also Wolfram Alpha series and coefficients),
where $\gamma$ is the Euler-Mascheroni constant, $$x=\dfrac{2-\sqrt\pi}{2+\sqrt\pi}\approx0.060317809,\quad \gamma\approx0.577215665.\tag4$$
The given inequality can be written in the form of $$\dfrac1{\sqrt\pi}\Gamma(y) < \dfrac{\pi\sqrt\pi}6\approx0.928054666,\tag5$$ wherein $$\dfrac1{\sqrt\pi}\Gamma(y) < 1 + \psi\left(\frac12\right)\left(y - \dfrac12\right) + \frac14\left(\pi^2+\psi^2\left(\frac12\right)\right)\left(y - \dfrac12\right)^2,\tag6$$ (see also Wolfram alpha series and coefficients),
where $\psi(x)$ is digamma function, $$\psi\left(\frac12\right) = -\gamma - 2\ln2 \approx-1.963510026.\tag7$$
From $(3),(4)$ should $$y-\dfrac12 < \frac1{25}.\tag8$$
From $(6)-(8)$ should $$\dfrac1{\sqrt\pi}\Gamma(y) < 1 + \dfrac1{25}\psi\left(\frac12\right) + \dfrac1{2500}\left(\pi^2+\psi^2\left(\frac12\right)\right)\approx0.926949589 < \dfrac{\pi\sqrt\pi}6,$$ i.e. inequality $(5)$ is correct.
Therefore, $$\color{brown}{\mathbf{\Gamma\left(\frac12\Gamma\left(\frac12\Gamma\left(\frac12 \right)\right)\right) < \frac{\pi^2}6}}.$$