Why does FS&K's SHAd-256 construction zero-prefix the inner message?
If $H(x)$ is a random oracle, then $H(H(0^b \mathbin\| x))$ is indifferentiable from a random oracle[1] (Theorem 5, p. 444; note that [1] uses the name ‘HMAC’ for this particular construction, at odds with the standard meaning of the name), but $H(H(x)) = H^2(x)$ is not[2], and there are protocols that can be exhibited (as [2] does) which are secure if instantiated with $H$ but trivially breakable if instantiated with $H^2$.

The reason that $H(H(0^b \mathbin\| x))$ works as a random oracle is essentially the domain separation of the inner hash and the outer hash: there is no string of the form $0^b \mathbin\| x$ that coincides with a string that $H(\cdot)$ could yield as an intermediate value.

The protocol in [2] that is breakable with $H^2$ but not $H$ is a mutual proof-of-work game:  Two parties aim to prove to each other that they have each spent the work to compute $\hat H^n(x)$ and $\hat H^m(y)$ for some $(n, x)$ and $(m, y)$ chosen by the opposite party, and for some hash $\hat H$.  When instantiated with $\hat H = H$, the protocol works; when instantiated with $\hat H = H^2$, one party can ask the other to compute $u = \hat H^n(x)$ where $x = H(y)$ and $n = m - 1$, and then return $$H(u) = H(\hat H^n(x)) = H(\hat H^n(H(y))) = \hat H^{n + 1}(y) = \hat H^m(y)$$ at a total cost of two $H$ evaluations, rather than $m$ as expected, completely destroying the protocol.

This protocol is contrived.  Maybe your protocol doesn't have this affliction.  But while there is a theorem that $H(H(0^b \mathbin\| x))$ is safe for all protocols, this counterexample means that you have to separately study any protocol that you consider instantiating with $H(H(x))$ to see whether $H(H(x))$ is a problem for that protocol.

The theorem serves as a free cryptographer for everyone to sign off on the use of $H(H(0^b \mathbin\| x))$; if you want to use $H(H(x))$ you have to hire a personal cryptographer of your own to review your particular use of $H(H(x))$, and cryptographers aren't cheap.  If you already have a personal cryptographer (or auditor), the theorem lets them save work auditing your work and spend their time and energy on the rest of the work.