Let $g(x)=2 f\left(\frac{x}{2}\right)+f(2-x)$ and f''(x) < 0 for all $x \in(0,2)$. Then, g(x), is 

increasing on $(0,4 / 3)$ and decreasing on $(4/3,2)$

We have,

$g(x) =2 f\left(\frac{x}{2}\right)+f(2-x)$ for all $x \in(0,2)$

$\Rightarrow g'(x)=f'\left(\frac{x}{2}\right)-f'(2-x)$ for all $x \in(0,2)$

Now,

$f''(x)<0$ for all $x \in(0,2)$

⇒ f'(x) is decreasing on (0, 2)

$\Rightarrow f'\left(\frac{x}{2}\right)>f'(2-x)$, if $\frac{x}{2}<2-x$

and,

$f'\left(\frac{x}{2}\right)<f'(2-x)$, if $\frac{x}{2}>2-x$

$\Rightarrow f'\left(\frac{x}{2}\right)>f'(2-x)$, if $x<\frac{4}{3}$ and, $f'\left(\frac{x}{2}\right)<f'(2-x)$, if $x>\frac{4}{3}$

$\Rightarrow g'(x)>0$, if $x \in(0,4 / 3)$ and, $g'(x)<0$, if $x \in(4 / 3,2)$

$\Rightarrow g(x)$ is increasing on $(0,4 / 3)$ and decreasing on $(4 / 3,2)$