Transformation behavior of connection on vector bundle. Using the notation from Jost's various books on geometry, let
$$
D=d+A
$$
be a connection on a vector bundle $\pi:E\rightarrow M$ with structure group $GL(n,\mathbb{R})$.  Also let $\{U_\alpha\}$ be an open covering for $M$ that yields local trivialisations with transition maps
$$
\varphi_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow GL(n,\mathbb{R}).
$$
Then $D$ defines a $T^*M$-valued matrix $A_\alpha$ on $U_\alpha$.  Let a section $s$ be given locally on $U_\alpha$ by $s_\alpha=s^i_\alpha\mu_i$, where $\{\mu_1,...,\mu_n\}$ is a frame for $E_{|_U}=\pi^{-1}(U)$.
Question I:  Why does it hold that
$$
s_\beta=\varphi_{\beta\alpha}s_\alpha\qquad\text{on $U_\alpha\cap U_\beta$}?
$$
Question II:  Why does it follow that
$$
\varphi_{\beta\alpha}(d+A_\alpha)s_\alpha=(d+A_\beta)s_\beta\qquad\text{on $U_\alpha\cap U_\beta$}?
$$
(He does give an "indication" of how this holds, but I don't see what he means.)
Question III:  How do we then conclude that
$$
A_\alpha=\varphi_{\beta\alpha}^{-1}d\varphi_{\beta\alpha}+\varphi_{\beta\alpha}^{-1}A_\beta\varphi_{\beta\alpha}?
$$
Remark:  Jost states this in each of his books on geometry, but I have never been able to find an elaboration.  I would also be grateful for some other references that explain in more detail what is going on here.
 A: The following is taken more or less directly from section 4.1 of Jost's Riemannian Geometry and Geometric Analysis. (Where he uses $\mu$, I will use $s$ to match the OP's notation.)
I think you may be getting confused when you try to introduce the frame $\{ \mu_i \}$. I don't think there's any need to mention frames here.
Consider a section $s$ of $E$. On each $U_\alpha \subset M$ over which $E$ is trivial, we can represent $s$ by a local section $s_\alpha$. This means that $s_\alpha$ is a map $U_\alpha \to \mathbb{R}^n$, i.e., a vector-valued function on $U_\alpha$. $\varphi_{\beta \alpha}$ is a map $U_\alpha \cap U_\beta \to Gl(n, \mathbb{R})$, i.e., for each point $p \in U_\alpha \cap U_\beta$, $\varphi_{ \beta \alpha}(p)$ is an invertible linear map $\mathbb{R}^n \to \mathbb{R}^n$. $s_\alpha$ and $s_\beta$ are related, for $p \in U_\alpha \cap U_\beta$, by $s_\beta (p) = [\varphi_{\beta \alpha}(p)] ( s_\alpha(p))$ (to answer your Question I, this is essentially just the definition of the transition map $\varphi_{\beta \alpha}$). (I will drop the reference to the point $p$ from now on.)
Now, on to the connection $D$. As Jost discusses and as you mention, $D$ defines locally on $U_{\alpha}$ a matrix $A_\alpha$ with one-form entries. 
Again, we use the local trivialization over $U_\alpha$ to view our section $s$ locally as a map $s_\alpha: U_\alpha \to \mathbb{R}^n$. $D s$ is a section of $E \otimes T^\ast M$, meaning locally $Ds$ is a sum of terms of the form $\sigma \otimes \omega$, where $\sigma$ is a section of $E$ and $\omega$ is a one-form. If we write the "$E$-piece" of $Ds$ locally in terms of the local trivialization of $E$, we can view $Ds$ locally as a vector with one-form entries that I'll call $(Ds)_\alpha$ (my notation, not Jost's). As Jost discusses, $(Ds)_\alpha = ds_\alpha + A_\alpha s_\alpha$, where $d$ is the usual exterior derivative acting componentwise on the entries of the vector-valued function $s_\alpha$, and $A_\alpha$ acts on $s_\alpha$ by matrix multiplication to give a vector with one-form entries.
Now to your Question II. On $U_\alpha \cap U_\beta$, we can write $Ds$ as a vector with one-form entries in two different ways corresponding to the two local trivializations: $(Ds)_\alpha$ and $(Ds)_\beta$. These two ways should be compatible in the sense that applying $\varphi_{\beta \alpha}$ to $(Ds)_\alpha$ should give us $(Ds)_\beta$, and this is where the equation you wrote comes from:
\begin{align*}
  \varphi_{\beta \alpha} ((Ds)_\alpha) &= (Ds)_\beta, \text{ i.e.,} \\
  \varphi_{\beta \alpha} ((d+ A_\alpha) s_\alpha) &= (d+ A_\beta) s_\beta
\end{align*}
Finally, we substitute the fact that $s_\beta = \varphi_{\beta \alpha}  s_\alpha$ into the above equation to solve for $A_\alpha$ in terms of $A_\beta$. This answers your Question III:
\begin{align*}
  \varphi_{\beta \alpha} ((d+ A_\alpha) s_\alpha) &= (d+ A_\beta) (\varphi_{\beta \alpha}  s_\alpha) \\
  \varphi_{\beta \alpha} (d s_\alpha) + \varphi_{\beta \alpha} (A_\alpha s_\alpha) &= (d\varphi_{\beta \alpha})s_\alpha + \varphi_{\beta \alpha} (ds_\alpha) +  A_\beta \varphi_{\beta \alpha}  s_\alpha \text{ (using Leibniz rule for $d$)}\\
 \varphi_{\beta \alpha} (A_\alpha s_\alpha) &= (d\varphi_{\beta \alpha})s_\alpha  +  A_\beta \varphi_{\beta \alpha}  s_\alpha \\
 A_\alpha s_\alpha &=  \varphi_{\beta \alpha}^{-1} (d\varphi_{\beta \alpha} +  A_\beta \varphi_{\beta \alpha} ) s_\alpha \\
\end{align*}
This holds for every $s_\alpha$ (the section $s$ was arbitrary), so we have the following equality of matrices with one-form entries on $U_\alpha \cap U_\beta$:
$$ A_\alpha =  \varphi_{\beta \alpha}^{-1} (d\varphi_{\beta \alpha} +  A_\beta \varphi_{\beta \alpha} ) $$
