Show a compound function is increasing and supermodular given its components
I don't think this proposition holds without assuming that $g$ is also increasing.
Take $f(x_1,x_2)=x_1x_2$ (actually any supermodular function) and $g(z)=-z$ (decreasing and convex). Then, $g(f(x_1,x_2))=-x_1x_2$ which is neither increasing in its arguments or supermodular.

If you assume also that $g$ is increasing, you get $g'(f(x))\geq 0$ straightforwardly.

You can easily show supermodularity using the conddition on the second derivatives. Denote $x=(x_1,...,x_n)$. Supermodularity is equivalent for function f to $$\frac{\partial^2 f}{\partial x_i \partial x_j}(x)\geq 0$$ for any $i\ne j$.
If you twice-differentiate $g \otimes f$ ($\otimes$ is the compound operator),
$$
\frac{\partial^2 g \otimes f}{\partial x_i \partial x_j}(x)=\frac{\partial^2 f}{\partial x_i \partial x_j}(x) . g'(f(x)) + \frac{\partial f}{\partial x_i}(x) .\frac{\partial f}{\partial x_j}(x) .g''(f(x))
$$
You can check that $\frac{\partial^2 g \otimes f}{\partial x_i \partial x_j}(x)\geq 0$ so that $g \otimes f$ is supermodular since $f$ is supermodular, $g$ is increasing and $g$ is convex.