Derivation for 6.109
We start with Equation 6.106
\[\begin{equation*}
\frac{v_j^{n+1} - v_j^{n-1} }{2 \Delta t} = \frac{1}{\rho_j\,N\Delta x} \sum_{\ell=0}^{N-1} i \ell \Delta k \, \tilde{\sigma}^n_\ell\, \exp{(2\pi i \ell j / N)} \quad.
\end{equation*}\]
Let us now now consider the relationship of \(v_j^{n+1}\) and \(v_j^{n-1}\) with their counterparts in the wavenumber domain. From 6.104 we have
\[\begin{align*}
v_j^{n+1} &= \frac{1}{N \Delta x} \sum_{\ell=0}^{N-1} \tilde{v}^{n+1}_\ell\, \exp{(2\pi i \ell j / N)}, \\
v_j^{n-1} &= \frac{1}{N \Delta x} \sum_{\ell=0}^{N-1} \tilde{v}^{n-1}_\ell\, \exp{(2\pi i \ell j / N)},
\end{align*}\]
such that Equation 6.106 may be written as
\[\begin{equation*}
\frac{\frac{1}{N \Delta x} \sum_{\ell=0}^{N-1} \tilde{v}^{n+1}_\ell\, \exp{(2\pi i \ell j / N)} - \frac{1}{N \Delta x} \sum_{\ell=0}^{N-1} \tilde{v}^{n-1}_\ell\, \exp{(2\pi i \ell j / N)}}{2 \Delta t} = \frac{1}{\rho_j\,N\Delta x} \sum_{\ell=0}^{N-1} i \ell \Delta k \, \tilde{\sigma}^n_\ell\, \exp{(2\pi i \ell j / N)} \quad.
\end{equation*}\]
This can be simplified to
\[\begin{equation*}
\frac{1}{2 \Delta t}\sum_{\ell=0}^{N-1} \left( \tilde{v}^{n+1}_\ell\, - \tilde{v}^{n-1}_\ell \right) \exp{(2\pi i \ell j / N)} = \frac{1}{\rho_j} \sum_{\ell=0}^{N-1} i \ell \Delta k \, \tilde{\sigma}^n_\ell\, \exp{(2\pi i \ell j / N)} \quad,
\end{equation*}\]
where the \(\frac{1}{N\Delta x}\) cancels from both sides. Ultimately this is saying that the two discrete Fourier transforms are equal to one another. Due to the fact that the \(\exp{(2\pi i \ell j / N)}\) form an orthogonal basis for the different values of \(\ell\), we know that the Fourier coefficient (amplitude) of each term must be equal. That is to say, for each individual value of \(\ell\) the equation
\[\begin{equation*}
\frac{1}{2 \Delta t} \left( \tilde{v}^{n+1}_\ell\, - \tilde{v}^{n-1}_\ell \right) \exp{(2\pi i \ell j / N)} = \frac{1}{\rho_j} i \ell \Delta k \, \tilde{\sigma}^n_\ell\, \exp{(2\pi i \ell j / N)} \quad
\end{equation*}\]
must hold. This can be simplified to
\[\begin{equation*}
\frac{1}{2 \Delta t} \left( \tilde{v}^{n+1}_\ell\, - \tilde{v}^{n-1}_\ell \right) = \frac{1}{\rho} i \ell \Delta k \, \tilde{\sigma}^n_\ell \quad,
\end{equation*}\]
which is the equation we seek to obtain. Note here that, as the question suggests, we have assumed a homogeneous medium such that \(\rho_j =\rho\) everywhere.
Derivation for 6.110
The derivation for the second relationship is essentially identical.
We start with Equation 6.107
\[\begin{equation*}
\frac{\sigma_j^{n+1} - \sigma_j^{n-1} }{2 \Delta t} = \frac{1}{\mu_j\,N\Delta x} \sum_{\ell=0}^{N-1} i \ell \Delta k \, \tilde{v}^n_\ell\, \exp{(2\pi i \ell j / N)} \quad.
\end{equation*}\]
Next we take the wavenumber versions of the stress,
\[\begin{align*}
\sigma_j^{n+1} = \frac{1}{N \Delta x} \sum_{\ell=0}^{N-1} \tilde{\sigma}^{n+1}_\ell\, \exp{(2\pi i \ell j / N)}, \\
\sigma_j^{n-1} = \frac{1}{N \Delta x} \sum_{\ell=0}^{N-1} \tilde{\sigma}^{n-1}_\ell\, \exp{(2\pi i \ell j / N)},
\end{align*}\]
such that Equation 6.107 may be written as
\[\begin{equation*}
\frac{\frac{1}{N \Delta x} \sum_{\ell=0}^{N-1} \tilde{\sigma}^{n+1}_\ell\, \exp{(2\pi i \ell j / N)} - \frac{1}{N \Delta x} \sum_{\ell=0}^{N-1} \tilde{\sigma}^{n-1}_\ell\, \exp{(2\pi i \ell j / N)}}{2 \Delta t} = \frac{\mu_j}{\,N\Delta x} \sum_{\ell=0}^{N-1} i \ell \Delta k \, \tilde{v}^n_\ell\, \exp{(2\pi i \ell j / N)} \quad.
\end{equation*}\]
This can be simplified to
\[\begin{equation*}
\frac{1}{2 \Delta t}\sum_{\ell=0}^{N-1} \left( \tilde{\sigma}^{n+1}_\ell\, - \tilde{\sigma}^{n-1}_\ell \right) \exp{(2\pi i \ell j / N)} = \mu_j \sum_{\ell=0}^{N-1} i \ell \Delta k \, \tilde{v}^n_\ell\, \exp{(2\pi i \ell j / N)} \quad,
\end{equation*}\]
where, under the same argument as before, we can reduce this to
\[\begin{equation*}
\frac{1}{2 \Delta t} \left( \tilde{\sigma}^{n+1}_\ell\, - \tilde{\sigma}^{n-1}_\ell \right) = \mu i \ell \Delta k \, \tilde{v}^n_\ell\quad.
\end{equation*}\]
