OK. First of all, change $(y_1,y_2)$ to $(y_1,\sqrt{B/A}y_2)$ and the time $z$ to $t=\alpha z+\frac{\Phi_A+\Phi_B}{2}$. Then we'll get the system with the matrix
$$
A(t)=\begin{pmatrix}
0 & C\cos (t + \Phi) \cr
C \cos(t- \Phi) & 0 \end{pmatrix}
$$
where $C=\alpha^{-1}\sqrt{AB}$ and $\Phi=(\Phi_B-\Phi_A)/2$. We want our approximation to be decent on $[\alpha z_{\min} ,\alpha z_{\max}]\subset [-\alpha,\alpha]$. Noting that $\cos(t-\Phi)=\cos(\Phi-t)$, we see that $A(t)=A^*(2\pi-t)$, which immediately tells us that the monodromy matrix $M$ from $0$ to $2\pi$ is self-adjoint. Also, denoting $\psi(t)=C\cos(\Phi+t)$=ce^{it}+\bar c{e^{-it}}$ with $c=\frac 12Ce^{i\Phi}, we see that the fundamental matrix $M(t)$ of the solution on $[0,2\pi]$ can be obtained (by the standard Piquard iterations) as the sum
$$
M(t)=\begin{pmatrix}
1 & 0 \cr
0 & 1 \end{pmatrix}+
\begin{pmatrix}
0 & \psi_1(t) \cr
-\psi_1(-t) & 0 \end{pmatrix}
$$
$$+
\begin{pmatrix}
-\psi_2(t) & 0 \cr
0 & -\psi_2(-t) \end{pmatrix}+
\begin{pmatrix}
0 & -\psi_3(t) \cr
\psi_3(-t) & 0 \end{pmatrix}+
O(C^4)
$$
Where $\psi_0(t)=1$ and $\psi_{k+1}(t)=\int_0^t\psi(s)\psi_k(-s)ds$.
You can write a long series but I want to convince you that 4 first terms are enough for your problem. We can find the first two $\psi$'s:
$$
\psi_1=\frac 1i[(ce^{it}-\bar c{e^{-it}})-(c-\bar c)
$$
$$
\psi_2=-\frac{c-\bar c}{i}\psi_1+\frac 1i(c^2-\bar c^2)t+\frac{|c|^2}2(e^{2it}+e^{-2it})-|c|^2
$$
and the linear term in $\psi_3$, which is
$$
(c^2-\bar c^2)t[(c-\bar c)+ce^{it}-\bar ce^{-it}]
$$
Plugging in $t=2\pi$, we see that the monodromy matrix is $\begin{pmatrix}
1-2\pi v & 4\pi vs \cr
4\pi vs & 1+2\pi v \end{pmatrix}+\begin{pmatrix}
O(C^4) & O(C^5) \cr
O(C^5) & O(C^4)\end{pmatrix}$
with $v=2\Im (c^2)=\frac 12 C^2\sin 2\Phi= \frac 12 \alpha^{-2}AB\sin(\Phi_B-\Phi_A)$, $s=c-\bar c=C\sin\Phi=\alpha^{-1}\sqrt{AB}\sin\frac{\Phi_B-\Phi_A}2$.
Now, the life is easy: the growth/decay part is essentially given by the matrix
$$
G(t)=\begin{pmatrix}
e^{-vt} & 0 \cr
0 & e^{vt} \end{pmatrix}
$$
the rotation part is essentially given by
$$
T=\begin{pmatrix}
1 & -s \cr
s & 1 \end{pmatrix}
$$
and the oscillation is essentially given by
$$
H(t)=\begin{pmatrix}
1-\widetilde\psi_2(t) & \psi_1(t) \cr
-\psi_1(-t) & 1-\widetilde\psi_2(-t) \end{pmatrix}
$$
where $\widetilde \psi_2$ is $\psi_2$ with the term $\frac 1i(c^2-\bar c^2)t$ removed.
So, $M(t)\approx H(t)T^{-1}G(t)T$ (well, $T^{-1}GT$ and $H$ don't commute but the commutator effect is of size $C^3$).
This should work just fine letting you to see just enough in your range.
P.S. My original answer had an error in that I neglected the rotation of eigenvectors, but now it should be fine even for fairly large $C$ like $0.1$. Check agaist your numerics and see if it works well enough for you.
