Q:

Why does this show that operator is well defined?

A function 
$g:\mathbb{R}^d\to\mathbb{R}^d,~\eta\to\alpha (|\eta |^2)\eta~$ with $\alpha\in C^1$ and strictly monotonically increasing is given. $g$ induces the composition operator $G:L^2(\Omega,\mathbb{R})\to L^2(\Omega,\mathbb{R}),~\eta\to\alpha (|\eta |^2)\eta$.
Now there is another operator $L:H_0^1(\Omega )\to (H_0^1(\Omega ))^*$ which is induced by $g$ (I guess in the same way like $G$).
The following inequality shows that $L$ is well defined:
$\langle L(v), w \rangle \leq \int_\Omega g(\nabla v(x), \nabla w(x))~dx = (G(\nabla v)|\nabla w)_{L^2} \leq c ||\nabla v||_2 ||\nabla u||_2$
But I do really not understand why this shows that $L$ is well defined..?

A:

These assumptions are not sufficient to show that $L$ is well-defined!
I take your post as that $L:H_0^1(\Omega) \to H_0^1(\Omega)^*$ is defined by
$$
(L(v))(w) =\int_\Omega \alpha( |\nabla v|^2) \nabla v\cdot \nabla w \ dx.
$$
So $L$ is well-defined, if this integral exists for all $v,w$.
Now take $\alpha(x)=x$. Then 
$$
(L(v))(v) =\int_\Omega |\nabla v|^2 \nabla v\cdot \nabla v \ dx = \|\nabla v\|_{L^4(\Omega)}^2.
$$
There is no reason to assume that this is finite for all $v\in H^1_0(\Omega)$.
In order to obtain well-defined $L$, you need to assume that $\alpha$ is bounded.

