= Solution
In <geodesic slicing>, $\alpha=1$ and $\beta^i=0$. The normal acceleration therefore vanishes:
$$
n^\rho\nabla_\rho n_\mu
=a_\mu=D_\mu\log\alpha=0.
$$
Hence each integral curve of $n^\mu$ is an affinely parametrized timelike geodesic.
On the initial surface $\hat t=0$, the diagonal Kruskal metric makes the unit normal point purely in the $\hat t$ direction. The observer starts at $\hat r=0$ with $\dot{\hat r}=0$, so its initial unit four-velocity equals that normal. The observer's geodesic and the normal integral curve solve the same geodesic initial-value problem. Uniqueness therefore gives
$$
\boxed{u^\mu=n^\mu}
$$
throughout their common domain.
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