How can initial conditions be taken into account to calculate a system's terminal value using the final value theorem or some other technique
Ah ha!  There's a lot of obfuscation in the problem as stated, plus a bit of over-emphasis on the final value theorem.

$$H(z) = \frac{z^{-1} \phi / (1 - \phi z^{-1})}{1 - z^{-1} \psi / (1 - \psi z^{-1})} \tag 0$$

OK, so far so good.

It arises from the following feedback system
$$Y(z)(1-z^{-1}) = z^{-1}F_{\psi}\cdot Y(z) + z^{-1}F_{\phi}\cdot X(z) \tag{1}$$
where $F_{\alpha}(z)$ is a 1st order high pass filter with parameter $\alpha$ of the form
$$F_{\alpha}(z) = \frac{\alpha (1 - z^{-1})}{1 - \alpha z^{-1}} \tag{2}$$

This was bothering me yesterday, but I didn't notice the essential problem.  The general form of the filter described in (2) is first-order, and the delay operators ($z^{-1}$) on the right side of (1) potentially add a second pole.
The original transfer function (0) is in a screwy form, but it appears to be second-order.  So the whole problem manages to worm its way through the pole-zero cancellation check by virtue of being stated in such a non-standard way.  This question deserves to be an extra-credit problem in a control systems class, or problem 10 out of 10 in a final by a particularly sadistic teacher, or perhaps a routine problem if the instructor believes in homework sets that no one will be able to fully solve.
The problem is that there's a pole-zero cancellation that snuck right by me and everyone else.  It's evident in the left-hand side of (1), where the derivative of $y_k$ is the subject of the equation.
So the reason that you can't solve this problem as stated using the final value theorem is that you cannot adequately represent the system using a transfer function.  There may be some way to save this within transfer function notation, but I just tried and failed at the first step, so I'm going to do it in state-space.
So, first, shine some light on this obfuscated* problem.  I'll do that by substituting (2) into (1):
$$Y(z)(1-z^{-1}) = 
     z^{-1}\frac{\psi(1-z^{-1})}{1 - \psi z^{-1}}Y(z) + 
     z^{-1}\frac{\phi(1-z^{-1})}{1 - \phi z^{-1}}X(z) \tag 3$$
Now the pole-zero cancellation becomes painfully evident: we're expressing a filter of the form $1 - z^{-1}$ on the left-hand side, but when we divide through by $1 - z^{-1}$, it no longer appears in the equation.
Oh joy.
So, drawing this out as a feedback loop (and changing from $z^{-1}$ to $z$ notation, because I'm a control systems guy):

Notation is going to get a bit awkward here.  Bear with me.  The transfer function in (2) can be realized as a state-space system (where $x$ is a state variable, not the system input):
$$\begin{aligned}
x_k = \alpha x_{k-1} + (\alpha - 1)u_k \\
y_k = -x_k + u_k
\end{aligned}
\tag 4
$$
So (awkwardly) noting that $x_{1,k}$ is the first element of the state vector $\mathbf{x}$ at time $k$, but is not the system input $x_k$, then the 'top' input to the summing junction is (4) with $\alpha = \phi$ and an extra delay:
$$\begin{aligned}
\begin{bmatrix}x_{1,k} \\ x_{2,k}\end{bmatrix} = 
\begin{bmatrix} \phi &amp; 0 \\ 0 &amp; 0\end{bmatrix} 
  \begin{bmatrix}x_{1,k-1} \\ x_{2,k-1}\end{bmatrix} + 
\begin{bmatrix}\phi - 1 \\ 1\end{bmatrix} x_k \\
u_{1,k} = x_{2, k-1} - x_{1,k-1}
\end{aligned}
\tag 5
$$
The output of the system, $y_k$, is itself just a state variable (oh joy -- not sarcastically this time).  So just let $$y_k = x_{3, k} \tag 6$$.
The filter in the feedback path is again described by (4), with $\alpha = \psi$ and an extra delay:
$$\begin{aligned}
x_{4,k} = \psi x_{4,k-1} + (\psi - 1)x_{3, k} \\
u_{2,k} = x_{4,k-1} - x_{3, k-1}
\end{aligned}
\tag 7
$$
Finally, the integrator in the forward portion of the loop is:
$$x_{2,k} = x_{2,k-1} + u_{1,k} - u_{2,k} \tag 8$$
This goes together into a fourth-order system:
$$\begin{aligned}
\mathbf{x}_k = \begin{bmatrix}
\phi &amp; 0 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 0 &amp; 0 \\
-1 &amp; 1 &amp; 0 &amp; -1 \\
0 &amp; 0 &amp; \psi - 1 &amp; \psi
\end{bmatrix}
\mathbf{x}_{k-1} + \begin{bmatrix}\phi - 1 \\ 1 \\ 0 \\ 0\end{bmatrix}x_k \\
y_k = \begin{bmatrix}0 &amp; 0 &amp; 1 &amp; 0\end{bmatrix}
\end{aligned}
\tag 9
$$
Recast that into $\mathbf{x}_k = A \mathbf{x}_{k-1} + B\,u_k,\,y_k = C\mathbf{x}_k$, and you can test the $A$ and $C$ matrices to see that it is observable, at least for some combinations of $\phi$ and $\psi$ (it is, I checked).  Since there's pole-zero cancellation, we can assume that it isn't controllable (you can check that for yourself).
Because it's observable, we can consider the system
$$\mathbf{x}_k = A \mathbf{x}_{k-1} + \mathbf{x}_0 \delta(k)$$
The response of this system to any given $\mathbf{x}_0$ is $$Y(z) = C(\mathbf{I}z - A)^{-1}\mathbf{x}_0$$  Now you can find the system response for each initial value in $\mathbf{x}$ using the final value theorem.
* I looked it up -- &quot;obfuscate&quot; literally means &quot;to darken, to put into shadow&quot;.  So shining a light on the thing really is the antonym of &quot;obfuscate&quot;.