Let $\Omega $ smooth and open and $f$ and $g$ with good conditions. We have to solve in $ H^1(\Omega )$ the problems $$(D):\begin{cases} -\Delta u=f&\Omega \\ u=g&\partial \Omega \end{cases}\quad \text{and}\quad (N):\begin{cases}-\Delta u=f&\Omega \\ \frac{\partial u}{\partial \nu}=g&\partial \Omega \end{cases}.$$
Why do we have to solve $(D)$ weakly in $H^1_g(\Omega )=\{v\in H^{1}(\Omega )\mid v|_{\partial \Omega }=g\}$ whereas to solve weakly $(N)$ we can stay in $H^1(\Omega )$ (and we don't have to consider a space as $H^1_*(\Omega )=\{v\in H^1(\Omega )\mid \partial _\nu v=g\}$.
If my question is not clear enough, let me know.
The choice of the functional space for weak formulations in an important and delicate question. Technically speaking we could formulate weak problems in various ways, but the issue is that we must choose the correct setting so that everything makes sense (for example, what does it mean that $v|_{\partial \Omega }=g$ in a Sobolev space?) and we can actually prove the existence of a (unique) solution and perhaps the equivalence with the initial problem. Let's start with problem $(D)$, a classical path would be to start first with the homogeneous problem, i.e:
$$ \begin{cases} -\Delta u=f&\Omega \\ u=0&\partial \Omega \end{cases} $$
It should be clear that the right space for a weak formulation for this problem should be $H^1_0(\Omega)$, since it is the space of $H^1$ functions which are zero on the boundary (in the trace sense) and hence incorporate the boundary condition. With this precise setting we can prove via Riesz representation theorem that the weak problem:
$$\text{find} \ \ u \in H^1_0(\Omega) : \ \int_{\Omega} \nabla u \nabla v \, dx = \int_{\Omega} fv \, dx \qquad \forall v \in H^1_0(\Omega) $$
has a unique solution.
The non-homogeneous problem can now be attached in the following way: if $g$ and $\partial \Omega $ are sufficiently regular, say $g \in H^{1/2}(\partial \Omega)$ and $\partial \Omega \in C^1 $, then there exists $u_g \in H^1(\Omega) : \gamma_0(u_g)=g$, this means that $u_g |_{\partial \Omega}=g $ in consistent way. Now we let $u_0=u-u_g$ and notice that $\gamma_0(u_0)=0$, hence we can formulate an homogeneous problem for $u_0$. If we can prove (in a similar fashion as before) that the latter has a unique solution, then $u=u_0 + u_g$ will be a solution for the non-homogeneous.
Back to the Neumann problem, we can stay in $H^1(\Omega)$ because we incorporate the boundary condition in our weak formulation (notice that for Dirichlet problems the datum is instead incorporated in the space itself). By the way, your question implies a delicate issue, suppose that we solved the weak formulation:
$$\text{find} \ \ u \in H^1(\Omega) : \ \int_{\Omega} \nabla u \nabla v \, dx = \int_{\Omega} fv \, dx + \int_{\partial \Omega} gv \, dx \qquad \forall v \in H^1(\Omega) \ \ (*) $$
Then are we sure that $u_{\nu}=g$ on the boundary? It depends on the regularity of the solution, for example if $u$ is only in $H^1$ we are not able to recover the boundary condition. You ask why we couldn't set $H^1_*(\Omega )=\{v\in H^1(\Omega )\mid \partial _\nu v=g\}$ as functional space, the answer is that for functions in $H^1(\Omega)$ the restriction of the normal derivative (i.e, the trace operator $\gamma_1$) is not well-defined, we would need $H^2$ for instance. Finally, $H^1$ really seems the right and "cheapest" setting for this Neumann problem because everything in $(*)$ makes sense, provived $g \in L^2$ for example. The so called problem of "regularity of weak solutions" (i.e, what is the best regularity a weak solution can achieve?) and the "backward equivalence" comes later on.