Let me first say that I think Tobias Kienzler has done a great job of discussing the intuition behind your question in going from finite to infinite dimensions.
I'll, instead, attempt to address the mathematical content of Jackson's statements. My basic claim will be that
Whether you are working in finite or infinite dimension, writing the Schrodinger equation in a specific basis only involves making definitions.
To see this clearly without having to worry about possible mathematical subtleties, let's first consider
Finite dimension
In this case, we can be certain that there exists an orthnormal basis $\{|n\rangle\}_{n=1, \dots N}$ for the Hilbert space $\mathcal H$. Now for any state $|\psi(t)\rangle$ we define the so-called matrix elements of the state and Hamiltonian as follows:
\begin{align}
\psi_n(t) = \langle n|\psi(t)\rangle, \qquad H_{mn} = \langle m|H|n\rangle
\end{align}
Now take the inner product of both sides of the Schrodinger equation with $\langle m|$, and use linearity of the inner product and derivative to write
\begin{align}
\langle m|\frac{d}{dt}|\psi(t)\rangle=\frac{d}{dt}\langle m|\psi(t)\rangle=\frac{d\psi_m}{dt}(t)
\end{align}
The fact that our basis is orthonormal tells us that we have the resolution of the indentity
\begin{align}
I = \sum_{m=1}^N|m\rangle\langle m|
\end{align}
So that after taking the inner product with $\langle m|$, the write hand side of Schrodinger's equation can be written as follows:
\begin{align}
\langle m|H|\psi(t)\rangle
= \sum_{m=1}^N\langle n|H|m\rangle\langle m|\psi(t)\rangle
= \sum_{m=1}^N H_{nm}\psi_m(t)
\end{align}
Equating putting this all together gives the Schrodinger equation in the $\{|n\rangle\}$ basis;
\begin{align}
\frac{d\psi_n}{dt}(t) = \sum_{m=1}^NH_{nm}\psi_m(t)
\end{align}
Infinite dimension
With an infinite number of dimensions, we can choose to write the Schrodinger equation either in a discrete (countable) basis for the Hilbert space $\mathcal H$, which always exists by the way since quantum mechanical Hilbert spaces all possess a countable, orthonormal basis, or we can choose a continuous "basis" like the position "basis" in which to write the equation. I put basis in quotes here because the position space wavefunctions are not actually elements of the Hilbert space since they are not square-integrable functions.
In the case of a countable orthonormal basis, the computation performed above for writing the Schodinger equation in a basis follows through in precisely the same way with the replacement of $N$ with $\infty$ everywhere.
In the case of the "basis" $\{|x\rangle\rangle_{x\in\mathbb R}$, the computation above carries through almost in the exact same way (as your question essentially shows), except the definitions we made in the beginning change slightly. In particular, we define functions $\psi:\mathbb R^2\to\mathbb C$ and $h:\mathbb R^2\to\mathbb C$ by
\begin{align}
\psi(x,t) = \langle x|\psi(t)\rangle, \qquad h(x,x') = \langle x|H|x'\rangle
\end{align}
Then the position space representation of the Schrodinger equation follows by taking the inner product of both sides of the equation with $\langle x|$ and using the resolution of the identity
\begin{align}
I = \int_{-\infty}^\infty dx'\, |x'\rangle\langle x'|
\end{align}
The only real mathematical subtleties you have to worry about in this case are exactly what sorts of objects the symbols $|x\rangle$ represent (since they are not in the Hilbert space) and in what sense one can write a resolution of the identity for such objects. But once you have taken care of these issues, the conversion of the Schrodinger equation into its expression in a particular "representation" is just a matter of making the appropriate definitions.
Each coherent state $|\alpha\rangle$ is, by definition, an eigenvector of the lowering operator $a_-$ with eigenvalue $\alpha$, namely
\begin{align}
a_-|\alpha\rangle = \alpha|\alpha\rangle.
\end{align}
If we assume that $|\alpha\rangle$ is normalized to one (i.e $\langle \alpha|\alpha\rangle=1$), as one usually does, then we find that the expectation value of $a_-$ in a coherent state is
\begin{align}
\langle \alpha|a_-|\alpha\rangle = \alpha\langle \alpha|\alpha\rangle = \alpha
\end{align}
On the other hand, the raising operator is the adjoint (aka hermitian-conjugate), of the lowering operator and vice versa
\begin{align}
(a_+)^\dagger = a_-
\end{align}
It follows that the expectation value of the raising operator in a coherent state is
\begin{align}
\langle \alpha|a_+|\alpha\rangle = \langle\alpha| (a_+)^\dagger|\alpha\rangle^* =\langle\alpha|a_-|\alpha\rangle^* = \alpha^*
\end{align}
Putting these together, we get the result you are looking for;
\begin{align}
\langle \alpha| x|\alpha\rangle = \sqrt{\frac{\hbar}{2m\omega}}(\langle\alpha|a_+|\alpha\rangle + \langle\alpha|a_-|\alpha\rangle) = \sqrt{\frac{\hbar}{2m\omega}}(\alpha^* + \alpha).
\end{align}
Best Answer
Hint :
In my opinion you must deal with the inverse problem : that of finding the propagator $\:K(\mathbf{x},t\;;\;\mathbf{x}_{0},t_{0})\:$ from the non-relativistic Schr$\ddot{\rm{o}}$dinger equation, so to understand why it is the transition probability amplitude for a particle present at position $\:\mathbf{x}_{0}\:$ on time $\:t_{0}\:$ to propagate at position $\:\mathbf{x}\:$ on time $\:t\:$ and how to use it for the solutions of the equation.
If by your efforts you could answer your question directly without dealing with the inverse problem then you would gain much on mathematics but lose much on physics.
In the following I sketch the main points of the inverse problem :
1. Let the non-relativistic Schr$\ddot{\rm{o}}$dinger equation \begin{equation} i\hbar\;\dfrac{\partial \psi(\mathbf{x},t)}{\partial t}=H(\mathbf{x},t)\;\psi(\mathbf{x},t) \tag{sk-01} \end{equation} where $\:H(\mathbf{x},t)\:$ the Hamiltonian, an hermitian operator.
2. Let $\:H\:$ be dependent only on the position coordinate $\:\mathbf{x}\:$ but not on time, that is $\:H=H(\mathbf{x})$.
3. Now we suppose that $\:H\:$ has (real) eigenvalues $\:E_{\jmath}\:$ with corresponding eigenfunctions $\:u_{\jmath}(\mathbf{x})\:$ \begin{equation} H(\mathbf{x})u_{\jmath}(\mathbf{x})=E_{\jmath}\;u_{\jmath}(\mathbf{x}) \qquad \jmath=1,2,\cdots \tag{sk-02} \end{equation} where the index $\:\jmath\:$ takes discrete or continuous values. For simplicity we suppose that $\:\jmath\:$ takes discrete values and that the set of eigenfunctions $\:\lbrace u_{\jmath} \rbrace, \jmath=1,2,...\:$ is a complete basis of the Hilbert space of states $\:\psi $. So, any state $\:\psi(\mathbf{x},t)\:$ can be expressed as \begin{equation} \psi(\mathbf{x},t)=\sum_{\jmath}a_{\jmath}(t) \; u_{\jmath}(\mathbf{x}) \tag{sk-03} \end{equation}
4. So \begin{equation} \sum_{\jmath}\left[ i\hbar\;\dfrac{da_{\jmath}(t)}{dt} -E_{\jmath}a_{\jmath}(t)\right] u_{\jmath}(\mathbf{x})=\mathbf{0} \tag{sk-04} \end{equation} and from the completeness of the basis $\:\lbrace u_{\jmath}\rbrace\:$ \begin{equation} a_{\jmath}(t)=a_{\jmath}(t_{0})\; \exp \left[ -i \frac{E_{\jmath}}{\hbar}(t-t_{0})\right] \qquad \jmath=1,2,\cdots \tag{sk-05} \end{equation}
5. The coefficients $\:a_{\jmath}(t_{0}) \:$ are determined from the initial state of the system \begin{equation} \psi_{0}(\mathbf{x}) \equiv \psi(\mathbf{x},t_{0})=\sum_{\jmath}a_{\jmath}(t_{0}) \; u_{\jmath}(\mathbf{x}) \tag{sk-06} \end{equation}
6. Under these conditions we have the following general solution of the Schr$\ddot{\rm{o}}$dinger equation (sk-01) \begin{equation} \psi(\mathbf{x},t)=\sum_{\jmath}a_{\jmath}(t_{0})\; \exp \left[ -i \frac{E_{\jmath}}{\hbar}(t-t_{0})\right] \; u_{\jmath}(\mathbf{x}) \tag{sk-07} \end{equation}
7. Now, equation (sk-01) is linear, that is, if to two initial states $\:\psi_{0}^{(1)}(\mathbf{x}) \:$ and $\:\psi_{0}^{(2)}(\mathbf{x}) \:$ at time $\: t_{0}\:$ there correspond the solutions $\:\psi^{(1)}(\mathbf{x},t) \:$ and $\:\psi^{(2)}(\mathbf{x},t) \:$ respectively, then to the initial state $\:\psi_{0}=a_{1}\psi_{0}^{(1)}+a_{2}\psi_{0}^{(2)}\:$, where $\:a_{1},a_{2}\:$ are complex numbers, there corresponds the solution $\:\psi=a_{1}\psi^{(1)}+a_{2}\psi^{(2)}$. Any initial state $\:\psi_{0}(\mathbf{x})\:$ may be considered as the superposition (linear combination) of impulses expressed via the Dirac function $\:\delta^{3}(\mathbf{x})\:$ as \begin{equation} \psi_{0}(\mathbf{x}) = \int \limits_{\mathbf{x}_{0}} \delta^{3}(\mathbf{x}_{0}-\mathbf{x}) \;\psi_{0}(\mathbf{x}_{0}) \; d^{3}\mathbf{x}_{0} \tag{sk-08} \end{equation}
8. If our system is a single point particle and this particle is appeared suddenly at point $\:\mathbf{x}_{0}\:$ on time $\: t_{0}\:$, then the expression $\:\delta^{3}(\mathbf{x}_{0}-\mathbf{x})\:$ under the integral represents the state (wave) function of this particle. If to this initial state there corresponds the solution \begin{align} K(\mathbf{x},t \; \boldsymbol{;} \;\mathbf{x}_{0},t_{0})\equiv \: & \text{the state function at point $\:\mathbf{x}\:$ on time $\:t\:$} \nonumber\\ &\text{corresponding to the appearance of the} \nonumber\\ &\text{particle at point $\:\mathbf{x}_{0}\:$ on time $\:t_{0}\:$} \tag{sk-09} \end{align} then, because of the linearity discussed in the previous paragraph, the solution $\psi(\mathbf{x},t)$ corresponding to the initial state $\psi_{0}(\mathbf{x})$ of equation (sk-08) would be \begin{equation} \psi(\mathbf{x},t) = \int \limits_{\mathbf{x}_{0}}K(\mathbf{x},t \; \boldsymbol{;} \;\mathbf{x}_{0},t_{0}) \; \psi_{0}(\mathbf{x}_{0}) \; d^{3}\mathbf{x}_{0} \tag{sk-10} \end{equation}
9. In mathematical terms $\:K(\mathbf{x},t \; ; \;\mathbf{x}_{0},t_{0})\:$ is called kernel while in physical terms is called propagator since it is the transition probability amplitude for a particle present at position $\:\mathbf{x}_{0}\:$ on time $\:t_{0}\:$ to propagate at position $\:\mathbf{x}\:$ on time $\:t$.
10. We'll determine $\:K(\mathbf{x},t \; ; \;\mathbf{x}_{0},t_{0})\:$ from equation (sk-07) \begin{equation} K(\mathbf{x},t \; \boldsymbol{;} \;\mathbf{x}_{0},t_{0})=\sum_{\jmath}c_{\jmath}\; \exp \left[ -i \frac{E_{\jmath}}{\hbar}(t-t_{0})\right] u_{\jmath}(\mathbf{x}) \tag{sk-11} \end{equation} where $\:c_{\jmath}\:$ are the "coordinates" of the initial state $\:\delta^{3}(\mathbf{x}_{0}-\mathbf{x})\:$ relatively to the complete basis $\:\lbrace u_{\jmath} \rbrace, \jmath=1,2,...\:$ according to equation (sk-06)
\begin{equation} \delta^{3}(\mathbf{x}_{0}-\mathbf{x}) =\sum_{\jmath}c_{\jmath}u_{\jmath}(\mathbf{x}) \tag{sk-12} \end{equation}
11. Taking the inner product of above equation with $\:u_{k}(\mathbf{x})\:$ we have \begin{equation} \left\langle \delta^{3}(\mathbf{x}_{0}-\mathbf{x}),u_{k}(\mathbf{x})\right \rangle =\sum_{\jmath}c_{\jmath}\left\langle u_{\jmath}(\mathbf{x}),u_{k}(\mathbf{x})\right \rangle \tag{sk-13} \end{equation} The complete basis $\:\lbrace u_{\jmath} \rbrace, \jmath=1,2,...\:$ is orthogonal since its members are eigenfunctions of the hermitian operator $\:H$. Without loss of generality we suppose that these eigenfunctions are normalized (unit norm) so that
\begin{equation} \left\langle u_{\jmath}(\mathbf{x}),u_{k}(\mathbf{x})\right \rangle = \int \limits_{\mathbf{x}} u_{\jmath}(\mathbf{x}) \;u_{k}^{\boldsymbol{*}}(\mathbf{x})\;d^{3}\mathbf{x}=\delta_{\jmath k} \tag{sk-14} \end{equation} By equations (sk-14), (sk-13) \begin{equation} c_{k}=\left\langle \delta^{3}(\mathbf{x}_{0}-\mathbf{x}),u_{k}(\mathbf{x})\right \rangle = \int \limits_{\mathbf{x}} \delta^{3}(\mathbf{x}_{0}-\mathbf{x}) \;u_{k}^{\boldsymbol{*}}(\mathbf{x})\;d^{3}\mathbf{x}=u_{k}^{\boldsymbol{*}}(\mathbf{x}_{0}) \tag{sk-15} \end{equation} that is \begin{equation} c_{\jmath}=u_{\jmath}^{\boldsymbol{*}}(\mathbf{x}_{0}) \tag{sk-16} \end{equation}
so equations (sk-12) and (sk-11) yield respectively \begin{equation} \delta^{3}(\mathbf{x}_{0}-\mathbf{x}) =\sum_{\jmath}u_{\jmath}^{\boldsymbol{*}}(\mathbf{x}_{0})\;u_{\jmath}(\mathbf{x}) \tag{sk-17} \end{equation} and \begin{equation} K(\mathbf{x},t \; \boldsymbol{;} \;\mathbf{x}_{0},t_{0})=\sum_{\jmath} u_{\jmath}^{\boldsymbol{*}}(\mathbf{x}_{0})\;\exp \left[ -i \frac{E_{\jmath}}{\hbar}(t-t_{0})\right] u_{\jmath}(\mathbf{x}) \tag{sk-18} \end{equation}