It is important to distinguish between three group actions that are named "Galilean":
-The Galilean transformation group of the Eucledian space (as an automorphism group).
-The Galilean transformation group of the classical phase space (whose Lie algebra constitute a Lie subalgebra of the Poisson algebra of the phase space). This is the classical action.
-The Galilean transformations of the wavefunctions (which are infinite dimensional irreducible representations). This is the quantum action.
Only the first group action is free from the central extension. Both classical and quantum actions include the central extension (which is sometimes called the Bargmann group).
Thus, the central extension is not purely quantum mechanical, however, it is true that most textbooks describe the central extension for the quantum case.
I'll explain first the quantum case, then I'll return to the classical case and compare oth cases to the Poincare group.
In quantum mechanics, a wavefunction in general is not a function on the configuration manifold, but rather a section of a complex line bundle over the phase space. In general the lift of a symmetry (an automorphism of the phase space) is an automorphism of the line bundle which is therefore a $\mathbb{C}$ extension of the automorphism of the base space. In the case of a unitary symmetry, this will be a $U(1)$ extension. Sometimes, this extension is trivial as in the case of the Poincare group.
Now, the central extensions of a Lie group $G$ are classified by the group cohomology group $H^2(G, U(1))$. In general, it is not trivial to compute these cohomology groups, but the case of the Galilean and Poincare groups can be heuristically understood as follows:
The application of the Galilean group action $\dot{q} \rightarrow \dot{q}+v$ to the non-relativistic action of a free particle: $S = \int_{t_1}^{t_2}\frac{m }{2}\dot{q}^2dt$, produces a total derivative leading to $S \rightarrow S + \frac{m}{2}v^2(t_2-t_1) + mvq(t_2) - mvq(t_1)$:
Now Since the propagator $G(t_1, t_2)$ transforms as $ exp(\frac{iS}{\hbar})$ and the inner product $\psi(t_1)^{\dagger} G(t_1, t_2) \psi(t_2)$ must be invariant, we get that the wavefunction must transform as:
$\psi(t,q) \rightarrow exp(\frac{i m}{2\hbar}(v^2 t+2vq) \psi(t,q) $
Now, no application of a smooth canonical transformation can romove the total derivative from the transformation law of the action, this is the indication that the central extension is non-trivial.
The case of the Poincare group is trivial. The relativistic free particle action is invariant under the action of the Poincare group, thus the transformation of the wavefunction doen not acquire additional phases and the group extension is trivial.
Classically, the phase space is $T^{*}R^3$ and the action of the boosts on the momenta is given by: $p \rightarrow p + mv$, thus the generators of the boosts must have the form
$K = mvq$, then the action is easily obtained using the Poisson brackets{q, p} = 1, and the Poisson bracket of a Boost and a translation is non-trivial {K, p} = m.
The reason that the Lie algebra action acquires the central extension in the classical case is that the action is Hamiltonian, thus realized by Hamiltonian vector fields and vector fields do not commute in general.
The Iwasawa decomposition of the Lorentz group provides the answer to your second question:
$SO^{+}(3,1) = SO(3) A N$
where $A$ is generated by the Boost $M_{01}$ and $N$ is the Abelian group generated by $M_{0j}+M_{1j}$, $j>1$. Now both subgroups $A$ and $N$ are homeomorphic as manifolds to $R$ and $R^2$ respectively.
To your third question: The limiting process which produces the Galilean group from the Poincare group is called the Wingne-Inonu contraction. This contraction produces the non-relativistic limit.
Its relation to quantum mechanics is that there is a notion of contarction of Lie groups unitary representations, however not a trivial one.
Update
In classical mechanics, observables are expressed as functions on the phase space. see for example chapter 3 of Ballentine's book for the explicit classical realization of the generators of the Galilean group.
This is a case where the full geometric quantization recepie can be carried out. See the following two articles for a review. (The full proof appears in page 95 of the second article. The technical computations are more readable in pages 8-9 of the first article).
The central extensions appear in the process of prequantization.
First please notice that the Hamiltonian vector fields $X_f$ corresponding to the Galilean Lie algebra generators close to the non-centrally extended algebra,
(because, the hamiltonian vector field of constant functions vanish).
However, the prequantized operators
$\hat{f} = f - i\hbar(X_f - \frac{i}{\hbar}i_{X_f} \theta)$, ($\theta$ is a symplectic potential whose exterior derivative equals the symplectic form) close to the centrally extended algebra because their action is isomorphic to the action of the Poisson algebra.
The prequantized operators are used as operators over the Hilbert space of the square integrble polarized sections, thus they provide a quantum realization of the centrallly extended Lie algebra.
Regarding your second question, the Wingne-Inonu contraction acts on the level of the abstract Lie algebra and not for its specific realizations.
A given realization is termed "Quantum", if it refers to a realization on a Hilbert space (in contrast to realization by means of Poisson brackets, which is the classical one).
Preliminary remarks.
As Danu writes in his comment, the physics of the other four generators has to do with spacetime translations, one for each spatial direction, and one for time. But how do we see this explicitly in the math behind the somewhat odd-looking presentation of the Poincare group and its Lie algebra that Hall discusses.
First, recall that any $d+1$-dimensional Lorentz transformation is a Linear transformation on $\mathbb R^{d+1}$, so it can be representation by multiplication by a $(d+1)\times(d+1)$ matrix $\Lambda$.
Second, and most crucially, recall that translations of $\mathbb R^{d+1}$ are not linear transformations; there is no way to write spacetime translation as multiplication by a matrix.
However, here's the really cool thing. If we embed a copy of $(d+1)$-dimensional spacetime into the vector space $\mathbb R^{d+2}$, namely into a space with one higher dimension, then we can implement translations as linear transformations. Here's how it works.
The main construction.
For each $x\in \mathbb R^{d+1}$, we associate an element of $\mathbb R^{d+2}$ as follows:
\begin{align}
x \mapsto \begin{pmatrix}
x \\
1 \\
\end{pmatrix}
\end{align}
Now, for each Lorentz transformation $\Lambda\in \mathrm O(d,1)$, and for each spacetime translation characterized by a vector $a\in\mathbb R^{d+1}$, we form the matrix
\begin{align}
\begin{pmatrix}
\Lambda & a \\
0 & 1 \\
\end{pmatrix} \tag{$\star$}
\end{align}
and we notice what this matrix does to the embedded copies of points in $\mathbb R^{d+1}$;
\begin{align}
\begin{pmatrix}
\Lambda & a \\
0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
1 \\
\end{pmatrix}
= \begin{pmatrix}
\Lambda x+a \\
1 \\
\end{pmatrix}
\end{align}
Whoah! That's really cool! What has happened here is that when we augment the dimension of spacetime by one with the embedding give above, and when we correspondingly embed Lorentz transformations and translations appropriately into square matrices of dimension $d+2$, then we actually do get a way of representing both Lorentz transformations and translations as linear transformations on $\mathbb R^{d+2}$ that act in precisely the correct way on the copy of Minkowski space embedded in $\mathbb R^{d+2}$!
In other words, the Poincare group in $d+1$ dimensions can be thought of as the set of all $(d+2)\times(d+2)$ matrices of the form $(\star)$ where $\Lambda\in \mathrm O(d,1)$ and $a\in\mathbb R^{d+1}$.
What about the Lie algebra?
A natural question then arises: what do the Lie algebra elements look like as matrices when we represent the Lie group elements this way? Well, I quick standard computation will show you that the Lie algebra of the Poincare group can, in this representation, be regarded as all matrices of the form
\begin{align}
\begin{pmatrix}
X & \epsilon \\
0 & 0 \\
\end{pmatrix} \tag{$\star$}
\end{align}
where $X\in\mathfrak{so}(d,1)$ and $\epsilon\in \mathbb R^{d+1}$, precisely as Hall indicates. But from the remarks above, we see clearly that the parameter $\epsilon$ precisely corresponds to the generators that span the subspace of the Poincare algebra that yield spacetime translations.
Best Answer
Unfortunately you can't just define the Poincare group to be that, because in the standard treatment it is defined a little bit differently. What you defined is actually the Lorentz group. The Poincare group contains translations as well.
The Lorentz group $O(1,3)$ is the group of all $\Lambda \in GL(4, \mathbb{R})$ such that
$$\Lambda^T \eta \Lambda = \eta,$$
with $\eta = \operatorname{diag}(-1,1,1,1)$. This can be seen as the group of all "changes of orthonormal frames in spacetime".
Remember that one orthonormal reference frame is a set of vectors $\{e_\mu\}$ such that $g(e_\mu,e_\nu)=\eta_{\mu\nu}$. In that sense, given two such frames, the change of frame that takes components in one of them to components in the other is given by these elements.
With this, the proper Lorentz group is $SO(1,3)$ which basically means that you pick all elements of $O(1,3)$ with determinant $+1$. The orthocronous part just means that if $\Lambda \in O(1,3)$ has $\Lambda^0_0 > 0$ then it preserves the sense of time-like vectors.
Some authors seems to include "by default" the orthocronous requirement in the group $SO(1,3)$ (see for example Analysis, Manifolds and Physics by Choquet-Bruhat, vol. 1, page 290). Others leave it separately, so that you end up with a group $SO^+(1,3)$, but this is a question of convention.
Now the Poincare group $P(1,3)$ (which I don't know any standard notation for) is the group of all Lorentz transformations together with all spacetime translations. In other words, we have:
$$P(1,3)= \{(a,\Lambda) : a\in \mathbb{R}^4, \Lambda \in SO(1,3)\}$$
together with the multiplication defined by
$$(a_1,\Lambda_1)\cdot (a_2,\Lambda_2)=(a_1+\Lambda_1 a_2, \Lambda_1\Lambda_2)$$
Think like that: while elements of $SO(1,3)$ relates orthonormal frames with coincident origins, elements of $P(1,3)$ also allows for the shift of origin as well.
The action of $SO(1,3)$ in the Minkowski vector space $\mathbb{R}^{1,3}$ (not to be confused with flat spacetime - this is actually the "model" for spacetime's tangent spaces, which just happens to be possible to identify with spacetime itself in the flat case) is given by usual matrix multiplication, i.e., given $ \Lambda \in SO(1,3)$ you have:
$$\Lambda \cdot v = \Lambda v, \quad \forall v\in \mathbb{R}^{1,3}.$$
On the other hand, the action of $P(1,3)$ on the Minkowski vector space $\mathbb{R}^{1,3}$ is characterized by the fact that given $(a,\Lambda)\in P(1,3)$ you have:
$$(a,\Lambda)\cdot v = a + \Lambda v, \quad \forall v\in \mathbb{R}^{1,3}.$$
I don't know if it helps, but people like to compare this to the case in $\mathbb{R}^3$ where you have the rotation group $SO(3)$ and the Euclidean group $E(3)$ comprising rotations in $SO(3)$ with translations in $\mathbb{R}^3$ thus forming the group of rigid motions. This could be seen as the analogous construction in spacetime.
EDIT: Regarding the semi-direct product construction mentioned in comments, recall that given groups $N,H$ with $\varphi : H\to \operatorname{Aut}(N)$ a homomorphism into the group of automorphisms of $N$, we can build the semi-direct product as the set $N\times H$ with the product:
$$(a,b)\cdot (c,d)=(a\varphi(b)(c), bd)$$
the resulting group is denoted $N\rtimes H$. In the particular case it is clear that we have this construction with $N = \mathbb{R}^{1,3}$, $H = SO(1,3)$ and $\varphi : SO(1,3)\to \operatorname{Aut}(\mathbb{R}^{1,3})$ given by $\varphi(\Lambda)(v) = \Lambda v$. Thus
$$P(1,3)=\mathbb{R}^{1,3}\rtimes SO(1,3)$$