This question (v1) asks many questions. Let us here make some general remarks, which OP hopefully will find useful.
Noether's theorem only needs infinitesimal transformations to work. Hence the important object is not the set $G$ of finite transformations, but rather the set $\mathfrak{g}$ of infinitesimal transformations.
In general, the set $\mathfrak{g}$ does not have to constitute a Lie algebra or even a Lie algebroid. The "Lie bracket" of two infinitesimal transformations might only close on-shell, i.e. modulo Euler-Lagrange equations. (This is known as an open algebra.)
A horizontal infinitesimal transformation $\delta x^{\mu}$ changes the spacetime point $x^{\mu}$, while a vertical infinitesimal transformation $\delta_0 \phi^{\alpha}(x)$ changes the fields $\phi^{\alpha}(x)$ without moving the spacetime point $x$. A general infinitesimal transformation is a combination of horizontal and vertical infinitesimal transformations.
A vertical infinitesimal transformation is typically of the form
$$\tag{1} \delta_0 \phi^{\alpha}(x)
~=~\varepsilon^a(x) ~Y^{\alpha}_a(\phi(x),\partial\phi(x),x)
+ d_{\mu}\varepsilon^a(x)~ Y^{\alpha, \mu}_a(\phi(x),\partial\phi(x),x),$$
where $\varepsilon^a(x)$ are infinitesimal transformation parameters, which are coordinates of a section $\varepsilon(x)$ in a vector bundle $E$ over spacetime.
To apply Noether's first theorem for a finite subspace of global$^1$ infinitesimal transformations, one identifies a finite-dimensional subspace of sections $\varepsilon_{(1)}(x)$,$\ldots,$ $\varepsilon_{(m)}(x)$, in $E$. Thus the global
infinitesimal transformations are of the form
$$\tag{2} \varepsilon(x)~=~ \sum_{r=1}^m \omega^{(r)}~\varepsilon_{(r)}(x), $$
where the parameters $\omega^{(1)}$, $\ldots$, $\omega^{(m)}$, are $x$-independent. In coordinates,
$$\tag{3} \varepsilon^a(x)~=~ \sum_{r=1}^m \omega^{(r)}~\varepsilon^a_{(r)}(x). $$
--
$^1$ A global (local) transformation refers in this physics context to an $x$-independent ($x$-dependent) transformation, respectively. What are $x$-independent are here really the $\omega^{(r)}$ parameters, not necessarily the basis elements $\varepsilon_{(r)}(x)$. Thus the notion of global transformations depends in principle on the choice of section basis $\varepsilon_{(1)}(x)$,$\ldots,$ $\varepsilon_{(m)}(x)$. [Local and global transformation in physics should not be confused with the mathematical notion of locally and globally defined objects. All transformations in this answer (local as well as global) are assumed to be globally defined on the entire spacetime. Locally defined transformations take us to the realm of gerbes.]
Schottenloher's chapter 3 and 4 are somewhat "non-linear" in that a lot of chapter 3 only makes sense after you've seen what chapter 4 does with it. In this particular case, sure, theorem 3.10 as written would apply to $G=\mathrm{SO}(3)$, and you would get some $E$ as a central extension. But this is a useless observation, because we don't have any machinery that would tell us something about this $E$! What you really want is that $E$ splits as $E\cong G\times \mathrm{U}(1)$ so that the lifting $S:E\to\mathrm{U}(\mathbb{H})$ reduces to a proper linear representation $S_2 : G\to\mathrm{U}(\mathbb{H})$.
Chapter 4 presents Bargmann's theorem as the main tool for this - the sequence splits for a $G$ that is simply-connected and has $H^2(\mathfrak{g},\mathbb{R}) = 0$. As the universal cover of $\mathrm{SO}(3)$, $\mathrm{SU}(2)$ is simply-connected and so Bargmann's theorem will apply to it, which is why Schottenloher is using it in this example.
You never need to worry about applying theorem 3.10 to groups that aren't simply connected: Every projective representation of a group lifts to a projective representation of its universal cover and in fact all projective representations of the cover are also projective representations of $G$ because the universal cover is just a central extension by the fundamental group $\pi_1(G)$, so we can always apply this theorem to the universal cover to have a chance of applying Bargmann's theorem afterwards.
If "the universal cover is just a central extension by the fundamental group" doesn't convince you because we're actually talking about central extensions by $\mathrm{U}(1)$, not discrete fundamental groups, the rest of this answer is for you.
Explicitly, let $\tilde{G}$ be the universal cover, then we have the projection
$$ \hat{\pi} : \mathrm{U}(1)\times \tilde{G} \to G$$
and when $G$ is compact and not the product of smaller Lie groups, $\pi_1(G)\cong \mathbb{Z}_m$ for some prime $m$. There is an embedding $\iota : \mathbb{Z}_m\to \mathrm{U}(1)$ mapping the elements of $\mathbb{Z}_m$ to the $m$-th roots of unity, and so we have a "diagonal" inclusion
$$ \mathbb{Z}_m\to \mathrm{U}(1)\times \tilde{G}, \gamma\mapsto (\iota(\gamma), \gamma)$$
Let's call the image of this inclusion $A$. Then clearly $A$ lies in the kernel of the projection
$$ \hat{\pi} :\mathrm{U}(1)\times \tilde{G}\to G, (x,g) \mapsto \pi(g)$$
where $\pi$ is the normal projection from the universal cover, and so we can turn the central extension
$$ 1\to \mathrm{U}(1)\times \pi_1(G) \to \mathrm{U}(1)\times \tilde{G} \overset{\hat{\pi}}{\to} G \to 1$$
into
$$ 1\to (\mathrm{U}(1)\times \pi_1(G))/A \cong \mathrm{U}(1) \to (\mathrm{U}(1)\times \tilde{G})/A \to G\to 1$$
where the thing in the middle is a "non-trivial" central extension of $G$ but only in the same way that $\tilde{G}$ is a non-trivial central extension of $G$, i.e. this doesn't have anything to do with the Lie algebra or its $H^2(\mathfrak{g},\mathbb{R})$ - on the level of the algebras, this extension is trivial.
Best Answer
Yes, of course this reduces to Bargmann's theorem: Bargmann's theorem says that when $H^2(\mathfrak{g},0) = 0$, then every projective representation of a simply-connected $G$ can be lifted a unitary representation of $G$. While the proof of Bargmann's theorem involves talking about central extensions by $\mathrm{U}(1)$, the statement of the theorem does not.
The theorem you quote here says the same: When $D = 1$ and $\tilde{G} = G$, i.e. $G$ is simply-connected, it says that the projective representation $\rho$ of $G$ can be lifted to the unitary representation $u$ of $G$.