= Solution
For a tangential <covector>, the projected <derivative> is $D_\nu W_\gamma=P^\beta{}_{\nu}P^\delta{}_{\gamma}\nabla_\beta W_\delta$. Its second <derivative> is
$$
D_\mu D_\nu W_\gamma
=P^\alpha{}_{\mu}P^\beta{}_{\nu}P^\delta{}_{\gamma}
\nabla_\alpha\left(P^\eta{}_{\beta}P^\kappa{}_{\delta}\nabla_\eta W_\kappa\right).
$$
Expanding by the <product rule>, the <derivative> of the first inner projector gives $K_{\mu\nu}n^\eta P^\kappa{}_{\gamma}\nabla_\eta W_\kappa$. This contribution cancels on antisymmetrizing $\mu,\nu$, because $K$ is symmetric.
The <derivative> of the second inner projector gives $K_{\mu\gamma}n^\kappa P^\eta{}_{\nu}\nabla_\eta W_\kappa$. Differentiating tangentiality, $n^\kappa W_\kappa=0$, rewrites it as
$$
K_{\mu\gamma}n^\kappa P^\eta{}_{\nu}\nabla_\eta W_\kappa
=-K_{\mu\gamma}K_\nu{}^\kappa W_\kappa.
$$
The remaining term is the projected four-dimensional curvature <commutator>. Using the curvature convention of the question,
$$
[\nabla_\alpha,\nabla_\beta]W_\delta
=W_\xi\,{}^{(4)}R^\xi{}_{\delta\beta\alpha},
$$
and replacing the tangential $W_\xi$ by $W_\lambda P^\lambda{}_{\xi}$, we obtain
$$
[D_\mu,D_\nu]W_\gamma
=W_\lambda\left[
P^\lambda{}_{\xi}P^\alpha{}_{\mu}P^\beta{}_{\nu}P^\delta{}_{\gamma}
{}^{(4)}R^\xi{}_{\delta\beta\alpha}
-K_{\mu\gamma}K_\nu{}^\lambda+K_{\nu\gamma}K_\mu{}^\lambda\right].
$$
Since this holds for every tangential <covector>, the <Gauss equation> follows:
$$
\boxed{{}^{(3)}R^\lambda{}_{\gamma\nu\mu}
=P^\lambda{}_{\xi}P^\alpha{}_{\mu}P^\beta{}_{\nu}P^\delta{}_{\gamma}
{}^{(4)}R^\xi{}_{\delta\beta\alpha}
-K_{\mu\gamma}K_\nu{}^\lambda+K_{\nu\gamma}K_\mu{}^\lambda.}
$$
The signs use a timelike unit normal and the specified <commutator> convention; reversing a curvature convention without changing the <commutator> would change the comparison.
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