Let me rephrase your question first to see if I understand what you are asking:
Suppose we are given a principal $G$-bundle $P \longrightarrow M$, a connection one-form $\omega$ on $P$, and an open neighborhood $U \subset M$ over which we have locally trivialized $P$ via local cross-sections $s$ and $s'$. Write $A$ and $A'$ for the local gauge fields for $\omega$ in $U$ with respect to the trivializations given by $s$ and $s'$, respectively. If
$$A' = g^{-1} A g$$
on all of $U$ for some $g \in G$, then what is the automorphism $f: P \longrightarrow P$ such that $s' = f^{-1} \circ s$?
Now that the question has been restated, let me tell you why (my interpretation of) your question is ill-posed.
The problem is that you only have given us local gauge fields $A$ and $A'$ over one open neighborhood $U \subset M$, while to define a global gauge transformation relating $A$ and $A'$, we need to know what the local gauge fields are on all open neighborhoods in some local trivialization of $P$, or in other words we need to know what $A$ and $A'$ are as elements of $\Omega^1_M(\mathrm{Ad}(P))$. Since we don't know what $A$ and $A'$ are globally, there is nothing we can do.
Now if $P$ is trivial, so that we can take $U = M$, we can indeed find the corresponding global gauge transformation $f$, since in this case we know $A$ and $A'$ on all of $M$.
Let me first describe a general way to relate global gauge transformations with local gauge transformations. Let $f: P \longrightarrow P$ be a given bundle automorphism, and suppose we have a local trivialization $\{(U_\alpha, \psi_\alpha)\}$ of $P$. We can think of these trivializations as given by local sections $s_\alpha: U_\alpha \longrightarrow \pi^{-1}(U_\alpha)$. For any $m \in U_\alpha$ and $p \in \pi^{-1}(m)$, there is a unique element $g_\alpha(p) \in G$ such that $p = s_\alpha(m).g_\alpha(p)$. Now define a map
$$\bar{\phi}_\alpha: \pi^{-1}(U_\alpha) \longrightarrow G,$$
$$\bar{\phi}_\alpha(p) = g_\alpha(f(p))g_\alpha(p)^{-1}.$$
One can easily check that $\bar{\phi}_\alpha(p.g) = \bar{\phi}_\alpha(p)$, so that $\bar{\phi}_\alpha$ is constant on the fibers of $P$ and hence descends to a map $\phi_\alpha: U_\alpha \longrightarrow G$. It can be verified that the $\phi_\alpha$ patch together to form a section $\phi \in \Omega^0_M(\mathrm{Ad}(P))$, and furthermore the map $f \mapsto \phi$ gives a bijective correspondence between the group of gauge transformations $\mathscr{G}$ and $\Omega^0_M(\mathrm{Ad}(P))$. We also have that if $\omega$ has gauge field $A \in \Omega^0_M(\mathrm{Ad}(P))$ and $f^\ast \omega$ has gauge field $A' \in \Omega^0_M(\mathrm{Ad}(P))$, then
$$A^\prime_\alpha = \phi_\alpha^{-1}A\phi_\alpha - \phi_\alpha^{-1}d\phi_\alpha.$$
Now let us return to the question at hand. In the case of the trivial bundle, there is only one $U_\alpha$ so we will drop the subscript. You are interested in the local gauge transformation $\phi(m) = g$ for some $g \in G$. You want to know what $f \in \mathscr{G}$ corresponds to this $\phi$. You can easily check that $f$ is given by left multiplication by $g$:
$$f(m, h) = (m, gh).$$
In the above formula for $f$ we are imagining $P$ as $M \times G$, i.e. using the standard trivialization of the trivial principal $G$-bundle. The formula changes if we use a different choice of trivialization. But in any case, you can still show that the local gauge transformation associated to this $f$ is $\phi \equiv g$, so that
\begin{align*}
A' & = \phi^{-1} A \phi - \phi^{-1} d\phi \\
& = g^{-1} A g - g^{-1} dg.
\end{align*}
I have computed that the 1-form $\omega:= p^*\omega' \in \Omega^1(Q\times H', \mathfrak{h'})$ descends to a form on $Q\times_{\alpha}H'$
I don't think this is true: firstly as Ted points out, the Lie group homomorphism $\alpha:H\to H'$ induces a Lie algebra homomorphism $\dot{\alpha} = T_e\alpha:\mathfrak{h}\to\mathfrak{h}'$. For $(q,h')\in Q\times H'$ and $\xi\in\mathfrak{h}$, the vector along the quotient direction
$$
(q\cdot \xi, -\dot\alpha(\xi)\cdot h')\in T_{(q,h')}(Q\times H')
$$
when contracted with $\omega = p^*\omega'$ takes the value $-\dot\alpha(\xi)$, which is non-zero for $\xi\notin \ker \dot\alpha$. However this vector descends to the zero vector on $Q\times_\alpha H'$.
I haven't worked through the details, but I think the correct form to consider would be
$$
\omega = \dot\alpha(p_1^*\omega_Q) + p_2^*\omega',
$$
where $p_1:Q\times H'\to Q$ and $p_2:Q\times H'\to H'$ denote the respective projections. It doesn't suffer from the above problem: applying it to the quotient direction $(q\cdot \xi, -\dot\alpha(\xi)\cdot h')$ yields 0, and so it should descend to a 1-form on $Q\times_\alpha H'$, which will be a connection.
Edit: I went back to work out the details of this, and it seems I was incorrect in my guess. The correct form should be
$$
\omega_{(q,h')} :=\mathrm{Ad}_{(h')^{-1}}\left(\dot{\alpha}(p_1^*\omega_Q )_{(q,h')}\right) + (p_2^*\omega_L')_{(q,h')}=\mathrm{Ad}_{(h')^{-1}}\left(\dot{\alpha}(p_1^*\omega_Q )_{(q,h')} + (p_2^*\omega_R')_{(q,h')}\right)
$$
Here $\omega_L' (\omega_R')$ is the left-(right-)invariant Maurer-Cartan form on $H'$. Letting $\pi_\alpha:Q\times H' \to Q\times_\alpha H'$ denote the quotient map,
$\omega$ vanishes on vectors of the form $(q\cdot\xi, -\dot\alpha(\xi)\cdot h')$ (i.e. vectors along the fibres of $\pi_\alpha$), and is invariant under the quotient action. So it reduces to a form $\omega_\alpha$ on $Q\times_\alpha H'$
$$
\omega = \pi_\alpha^*\omega_\alpha.
$$
The obvious right actions $R_\cdot$ and $\check{R}_\cdot$ of $H'$ on $Q\times H'$ and $Q\times_\alpha H'$ commute with $\pi_\alpha$,
$$
\pi_\alpha\circ R_{h'} = \check{R}_{h'}\circ \pi_\alpha,
$$
and it is straightforward to verify that $R_{k'}^*\omega = \mathrm{Ad}_{(k')^{-1}}\omega$ for all $k'\in H'$. It follows that
$$
R_{k'}^*\pi_\alpha^*\omega_\alpha = \mathrm{Ad}_{(k')^{-1}}\pi_\alpha^*\omega_\alpha \implies \pi_\alpha^*\check{R^*}_{k'}\omega_\alpha = \pi_\alpha^*(\mathrm{Ad}_{(k')^{-1}}\omega_\alpha).
$$
Since $\pi_\alpha$ is a submersion, this implies $\check{R^*}_{k'}\omega_\alpha = \mathrm{Ad}_{(k')^{-1}}\omega_\alpha$. Also $\omega_{(q,h)}((q,h)\cdot \xi) =\omega_{(q,h)}(0_q,h\cdot \xi) = \xi$ and the fact that $\pi_\alpha$ commutes with the right action implies the equivalent result for $\omega_\alpha$. So $\omega_\alpha$ is a connection form on $Q\times_\alpha H'$.
Best Answer
I'll try, but I'm a bit lazy for working through your notation, and use my own. I guess it should be close enough so you can work that out...
For (1.):
By definition a connection form is a differential form $\mathcal{A}\in\Omega^1(P,\mathfrak{g})$, which is both of type $\text{Ad}$, i.e. $(R_g)^*\mathcal{A}=\text{Ad}\circ\mathcal{A}$, and in a point $p\in P$ the map $\mathcal{A}\big|_p:T_pP\to\mathfrak{g}$ has a certain behavior on vertical tangent vectors. That is, by being of type $\text{Ad}$, a connection form is determined already on a whole fiber by only one single value in this fiber. On the other hand, the difference between two connection forms is also a $1$-form of type $\text{Ad}$ and vanishes on vertical tangent vectors (since both have the same prescribed values on these). Such $1$-forms with the latter property are called horizontal. Conversely, you can show that for any connection form $\mathcal{A}$ and any horizontal $1$-form $\omega$ of type $\text{Ad}$, also $\mathcal{A}+\omega$ is a connection form. In other words, the set of connection forms is an affine space with spatial structure of the space of horizontal $1$-forms of type $\text{Ad}$.
Now, on a trivial bundle $U\times G$ you have a canonical flat connection form, which essentially does only what it should on vertical tangent vectors, and nothing else. Up to some issues of identification, I think this should be more or less what you have labeled $\theta_{(L)}$, or $\theta_{(L)}^A$ in some basis of $\mathfrak{g}$. Moreover, if you have a local trivialization $\phi:\pi^{-1}(U)\to U\times G$, then you can pull back the distinguished connection form on $U\times G$ to $\pi^{-1}(U)$.
Hence, your local trivialization did distinguish a connection form on $\pi^{-1}(U)$. As we have seen above, any other connection form, restricted to $\pi^{-1}(U)$, is then our distinguished one, plus some horizontal $1$-form of type $\text{Ad}$ on $\pi^{-1}(U)$, call it $\eta$. The local trivialization of $\pi^{-1}(U)$, moreover, distinguishes a section $s$ in $\pi^{-1}(U)$ (the one you have mentioned), and a $1$-form of type $\text{Ad}$ is fully determined by its values on the range of this section (which contains one point in each fiber). Then, if you pull back this $1$-form through a section $s:U\to\pi^{-1}(U)$, you end up with a $1$-form $s^*\eta$ on $U$ with values in $\mathfrak{g}$ and your original $\eta$ may be fully recovered from only the values of $s^*\eta$, namely, more or less, by the second term of your formula for $\bar{\omega}_p$ (where you additionally expanded in the basis $T_A$).
Taken together, locally on some $U$ on which both the principal bundle admits a trivialization and the basis manifold admits a chart, your formula is just expanding an arbitrary connection form into the basis $\text{d}x^\mu$, the basis $T_A$ and the part (first term) with the distinguished connection form on $\pi^{-1}(U)$.
Does this help you?
For (2.):
I'm not sure how to answer this. To me it seems that these "every other books" concentrate on the non-trivial part of a connection form which I have discussed above, and just call this the connection form. This is quite sloppy as not always possible, but suitable for many applications. From your first sentence I conclude that you head up to physics applications. Note that, e.g. Minkowski space is contractible (just $\mathbb{R}^4$), and thus every vector bundle (and the associated frame bundles) over it is trivializable. Therefore, such a bundle admits a distinguished (global) connection form on the frame bundle and it suffices to regard any other connection form only by its deviation from the distinguished one. Also if you head to GR applications, a metric also distinguishes a connection form on the frame bundle and we have the same situation.
For (3.):
As I said, I assume you mean by $\omega$ the connection on an associated vector bundle (probably on the adjoint bundle, as you expand $\omega$ with in indices $A$. Is that right?). Also on a vector bundle you can show that the set of connections is an affine space, but now with the spatial structure of $1$-forms on the basis manifold, which take values in the endomorphism bundle of the given vector bundle. Moreover, you find a one-on-one association between connections on a vector bundle, and connection forms on its frame bundle. Given a local trivialisation of the former, you also have a local trivialization of the latter. Moreover, given a local trivialization of the vector bundle you can distinguish a certain connection/covariant derivative, which is essentially just the Cartan derivative $\text{d}$ (concretely acting in your formula by $\partial_\mu$), pulled back through the local trivialization. This connection is, under the above one-on-one association, mapped to the trivial flat connection form on the respective part of the frame bundle, if you do everything correctly (I know, that's a stupid thing to say...). Moreover, this one-on-one association maps the deviation from the respective distinguished connection/connection forms to each other (where the latter is first pulled back through the right section, above called $s^*\eta$ to reproduce your forms $\omega_\mu$ or $\omega_\mu^A$, resp.). That's why you find your $\omega_\mu^A$ again in this connection, as deviation of your given connection form the distinguished trivial connection acting by $\partial_\mu$.
Does this help you?