Pseudorandom zero-sharing: how does it work?
To your question:


About the HMAC example, maybe the hypothesis that $f^i_A(j)$ is not feasible. Because $f_A(x)$ servers as a role somelike base polynomial while $\psi$ servers as a random scalors of these base polynomials. 


$f_A(x)$ is constructed in this way: first, we know $t$ roots of $f_A(x)$, then we assign $f_A(0)=1$. Therefore, $f_A(x)$ is a $t$ degree polynomial whose constant term is 1. Besides, other points like $(j, f_A(j))$ are computed by interpolating. We cannot simply treat $f_A(j)$ as 1.


About reconstructing secret 0. Now that we know the secret is 0, why we need to concern about how to reconstruct the secret. I think maybe the real secret, things we want to hide from is coefficients of non-constant terms in $f_A(x)$, because we usually use PRZS protocol to randomize an existing polynomial whose constant term is a secret.




BTW, there is a simpler version of PRZS:

\begin{equation*}
   f(X) = \sum_{A\subset X\text{, }|A|=n-t}\sum_{j=1}^t \psi_{r_A}(a, j) \cdot X^j \cdot f_A(X)
\end{equation*}

Three observations here:


$f_A(X)$ is a polynomial whose degree is at most $t$ with random constant term.
$X^t \cdot f_A(X)$ is a polynomial whose degree is at most $2t$.
$f(X)$ is divisible by $X$ which means its constant term is 0. Beacause $X^1 \cdot f_A(X)$ contributes $f(X)$'s lowest degree.


Therefore, we come to the conclusion that $f(X)$ is a polynomial that $deg(f) \leq 2t$, and $f(0) = 0$.