As pointed out by FraSchelle, your first question (why we can replace by the Brillouin zone by a sphere when calculating winding numbers) has been asked (and answered) a few times. The same goes for your tag-on question of why we get a $\mathbb Z_2$ invariant in the case of extra symmetries. So I will focus on your middle question, which I find the most interesting:
If $\pi_3(S^2) = \mathbb Z$, why don't we have a topological phase of matter corresponding to that?
Well: we do :) It is called a Hopf insulator (due to the fact that the non-trivial maps $S^3 \to S^2$ are the so-called Hopf maps).
But it is not as interesting as a Chern insulator, because a Chern insulator cannot be connected to a product state without a phase transition (i.e. it has intrinsic topological order--at least for some of the definitions). The Hopf insulator, however, can be trivialized. You might wonder how that is possible, given it has a non-zero discrete index given by the winding number of the Brillouin zone over $S^2$. Indeed: isn't the whole point of such discrete invariants that they show we cannot trivialize the state without a phase transition? Well, one objection might be that such a winding number is only well-defined if you have a Brillouin zone, which means you have to presume translation invariance and no interactions. The same can be said about the winding number for a Chern insulator, but it turns out that in that case the topological invariant can be extended even to the case where translation invariance is broken and/or interactions are added. (One usual argument to make this plausible is that the winding number in the case of the Chern insulator is equivalent to the discrete Hall conductance, with the Hall conductance being well-defined even without translation invariance or with interactions. Of course to make that a justified statement, one has to prove that the Hall conductance, if non-zero, is quantized. I don't know a good argument for this, but one argument I could imagine goes like this: having a Hall conductance means that your effective electromagnetic response in your material is given by $S = k \int A d A$ (indeed: use the Euler-Lagrange equations to show that this gives a Hall conductance $\propto k$). If one then believes this action should also work on the quantum level, then it is a well-known fact that this action (called a Chern-Simons action) is only well-defined if $k$ is discrete [it has to do with the fact that under a gauge transformation, $\int A d A$ changes by a multiple of $2\pi$, so if we want $e^{iS}$ to be gauge-invariant, we need $k \propto$ an integer.])
For the Hopf insulator I do not know such an extension/argument. But does this mean that even if we assume translation invariance and no interactions that our topological phase is non-trivial? Well: yes and no. In a strict way, yes, because then we have our non-zero topological invariant. But in a physical way, not really. This is because $\pi_3(S^2)$ presumes our system only has two bands. Indeed the ``$S^2$'' arises from the fact that for every momentum $\boldsymbol k$ we have $H_{\boldsymbol k} = \boldsymbol n_{\boldsymbol k} \cdot \boldsymbol \sigma$ (where we can assume $|n_{\boldsymbol k}| = 1$ and so we get $S^2$). So this doesn't tell us what happens if we allow for extra bands. In fact, when people classify these phases, they check what happens when you add bands, because we don't want to call a phase topological if by adding a trivial band to our system, we can now connect our whole state to a trivial product state without a phase transition. But it turns out in the case of the Hopf insulator this is exactly the case. (See e.g. https://arxiv.org/abs/1307.7206 for more information.)
