The matrix, as you note, is of the form
$$
\pmatrix{A&0&B\\0&\sigma_2 &0\\C&0&F}.
$$
With the block permutation matrix
$$
P = \pmatrix{0&I_2&0\\I_2&0&0\\0&0&I_2},
$$
we find that this matrix is similar to the block-diagonal matrix
$$
PDP^T = \pmatrix{A&B&0\\C&F&0\\0&0&\sigma_2}.
$$
So, the eigenvalues of $D$ are equal to those of $\sigma_2$ (namely $\pm 1$) together with those of the matrix $M = \left[\begin{smallmatrix}A&B\\C&F\end{smallmatrix}\right]$. On the other hand, $M$ is the matrix
$$
M = \left[\begin{array}{c|cc|c}
1&0&0&0\\
\hline 0&0&1&0\\
0&1&0&0\\
\hline 0&0&0&-1\end{array}\right].
$$
As the partition lines indicate, this matrix is block-diagonal with diagonal blocks $1$, $\left[\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right]$, $-1$.
With that, you can conclude that the eigenvalues of $D$ are equal to $1$ and $-1$, both with multiplicity $3$.
For a more direct approach, the permutation matrix
$$
P = \pmatrix{
1&0&0&0&0&0\\
0&0&0&0&0&1\\
0&0&1&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&1&0\\
0&1&0&0&0&0}
$$
is such that
$$
PDP^T = \pmatrix{\sigma_3 &0&0\\0&\sigma_2&0\\0&0&\sigma_1}.
$$
To find the eigenvectors of $D$, begin by finding the eigenvectors of $PDP^T$. We can find these eigenvectors using the eigenvectors of the blocks $\sigma_1,\sigma_2,\sigma_3$.
For example, $\sigma_2$ has the eigenvector $v = (1,i)$ associated with eigenvalue $1$. It follows that $PDP^T$ has the "block-vector"
$$
\tilde v = \pmatrix{0_{2 \times 1}\\ v\\ 0_{2 \times 1}}
$$
as an eigenvector associated with the eigenvalue 1.
Once all eigenvectors are obtained in this fashion, note that for any eigenvector $x$ of $PDP^T$, $P^Tx$ will be an eigenvector of $D$ because
$$
D(P^Tx) = (DP^T)x = P^T(PDP^T)x = P^T(\lambda x) = \lambda P^T x.
$$
So, once the eigenvectors $v_1,\dots,v_6$ of $PDP^T$ are obtained, $P^Tv_1,\dots,P^Tv_6$ will be the eigenvectors of $D$.
Best Answer
This matrix can be written as the Kronecker product $M \otimes I_a$, where $M$ is the $r \times r$ circulant matrix $$ M = \pmatrix{2&-1&0&\cdots&0&-1\\ -1&2&\ddots & \ddots & \vdots & 0\\ 0 & \ddots & \ddots & \ddots & 0 & \vdots\\ \vdots &0&\ddots&\ddots&-1&0\\ 0&\vdots&\ddots &-1 & 2 & -1\\ -1 & 0 &\cdots & 0& -1 & 2}. $$ It follows that the eigenvalues of your matrix are simply the eigenvalues of $M$, each repeated with $a$-fold multiplicity.
The eigenvalues of the circulant matrix $M$ can be computed using the formula here to be $$ \lambda_k = 2 - e^{2 \pi i k/r} - e^{-2 \pi i k/r} = 2(1 - \cos(2 \pi k/r)), \quad k = 0,1,\dots,r-1. $$