= Solution
The projection $X\mapsto X_\parallel$ is tensorial. Although a covariant derivative is not $C^\infty$-linear in its second argument, the extra term in $\nabla_{X_\parallel}(fY_\parallel)$ is proportional to $Y_\parallel$, whose contraction with the normal vanishes. Thus $K(X,Y)$ is $C^\infty$-linear in both arguments and defines a $(0,2)$ tensor on the hypersurface.
Since $n_aY_\parallel^a=0$,
$$
-n_a\nabla_{X_\parallel}Y_\parallel^a
=(X_\parallel)^c(Y_\parallel)^d\nabla_cn_d.
$$
Therefore the <extrinsic curvature> is
$$
\boxed{K_{ab}=h_a{}^ch_b{}^d\nabla_cn_d}.
$$
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