There is an alternative. A function is quasi-concave if all super-level sets of it are convex. A super-level set for a function $z(x)$ is defined as $S_\alpha(z) = \{ x\, | \,z(x) > \alpha\}.$
Therefore we consider $S_\alpha(h) = \{ (x_1,x_2)\, | \,h(x_1,x_2) > \alpha\} = \{(x_1,x_2)\, |\, f(g^1(x_1)+g^2(x_2)) > \alpha\}$
and since $f$ is a non-decreasing function so we have $S_\alpha(h) = \{(x_1,x_2)\, |\, g^1(x_1)+g^2(x_2) > \alpha\}$ Since according to problem statement, $g^1$ and $g^2$ are concave functions, so sum of them is also concave which have the pleasant indication that $S_\alpha(h) = \{(x_1,x_2)\, |\, g^1(x_1)+g^2(x_2) > \alpha\}$ is a convex set and therefore $h$ is a quasi-concave function.