Questions need to be answered, so I guess I will shoot:

Reduce[k (2 (a'[x])^2 - a[x] a''[x])/(a[x]^2) > 0 && k > 0]

$a'(x)\in \mathbb{R}\land \left(\left(a(x)<0\land a''(x)>\frac{2 a'(x)^2}{a(x)}\land k>0\right)\lor \left(a(x)>0\land a''(x)<\frac{2 a'(x)^2}{a(x)}\land k>0\right)\right)$

which is the result that you expected.

Regarding your comment:

Is there any way I wouldn't have to repeat k>0 in every function but settle it beforehand?

You can do:

assum = k > 0;
Reduce[k (2 (a'[x])^2 - a[x] a''[x])/(a[x]^2) > 0 && assum]

Questions need to be answered, so I guess I will shoot:

Reduce[k (2 (a'[x])^2 - a[x] a''[x])/(a[x]^2) > 0 && k > 0]

$a'(x)\in \mathbb{R}\land \left(\left(a(x)<0\land a''(x)>\frac{2 a'(x)^2}{a(x)}\land k>0\right)\lor \left(a(x)>0\land a''(x)<\frac{2 a'(x)^2}{a(x)}\land k>0\right)\right)$

which is the result that you expected.

Regarding your comment:

Is there any way I wouldn't have to repeat k>0 in every function but settle it beforehand?

You can do:

assum = k > 0;
Reduce[k (2 (a'[x])^2 - a[x] a''[x])/(a[x]^2) > 0 && assum]