= Solution
Keep the unit-normal sign explicit: $n^an_a=\varepsilon\in\{+1,-1\}$. Since the hypersurface is spacelike its normal is timelike. Construct
$$
n_a=\pm\frac{\nabla_af}{\sqrt{|g^{bc}\nabla_bf\nabla_cf|}},\qquad h_{ab}=g_{ab}-\varepsilon n_an_b,\qquad h_a{}^b=\delta_a{}^b-\varepsilon n_an^b.
$$
Choose the sign of n for the desired time orientation. The pullback of h is the nondegenerate <induced metric>. The <second fundamental form> is
$$
\boxed{K_{ab}=h_a{}^ch_b{}^d\nabla_cn_d.}
$$
Its sign reverses when n reverses. It is tangential in both indices and symmetric: the antisymmetric part of the derivative of a normalized gradient vanishes after both tangential projections. Equivalently it is the projected Hessian of f divided by the normalizing factor. This is the normal-change definition of <extrinsic curvature>.
For tangent vector fields X,Y, let D be the induced <Levi-Civita connection>. Orthogonality to n gives the derivative decomposition
$$
\nabla_XY=D_XY-\varepsilon K(X,Y)n,\qquad g(\nabla_Xn,Y)=K(X,Y).
$$
Use the curvature convention of the preceding solution, $R(X,Y)Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z$. Substitute the decomposition twice and take its tangential component. Terms differentiated normally cancel, while the two normal components contribute the quadratic shape terms:
$$
P R^{(4)}(X,Y)Z=R^{(3)}(X,Y)Z-\varepsilon\{K(Y,Z)\nabla_Xn-K(X,Z)\nabla_Yn\}.
$$
Lowering the first curvature index gives the <Gauss equation for a nonnull hypersurface>
$$
\boxed{{}^{(3)}R_{abcd}=h_a{}^eh_b{}^fh_c{}^gh_d{}^k{}^{(4)}R_{efgk}+\varepsilon(K_{ac}K_{bd}-K_{ad}K_{bc}).}
$$
All four displayed indices on the left are tangential.
Contract with $h^{ac}h^{bd}$ and use $h^{ab}=g^{ab}-\varepsilon n^an^b$ for tangential contractions. The double-normal term vanishes by curvature antisymmetry and each mixed contraction is ambient <Ricci curvature>. With $K=h^{ab}K_{ab}$ this gives
$$
\boxed{{}^{(3)}R={}^{(4)}R-2\varepsilon\,{}^{(4)}R_{ab}n^an^b+\varepsilon(K^2-K_{ab}K^{ab}).}
$$
This also records which metric and curvature conventions enter the sign.
In vacuum without a cosmological constant, the ambient <Ricci tensor> and scalar vanish. A <totally umbilic hypersurface> has $K_{ab}=\lambda h_{ab}$, so $K=3\lambda$ and $K_{ab}K^{ab}=3\lambda^2$. Therefore the <scalar curvature of an umbilic vacuum hypersurface> is $6\varepsilon\lambda^2$.
To express the requested nonnegative version, take Lorentz signature $(+---)$, so the timelike <unit normal> has $\varepsilon=+1$ and h is the induced negative-definite spatial metric. Then
$$
\boxed{{}^{(3)}R=6\lambda^2=\frac23K^2\geq0.}
$$
If instead one uses signature $(-+++)$ and its positive-definite spatial metric with the same curvature-slot convention, the scalar is $-6\lambda^2$. Replacing h by $-h$ reverses its scalar curvature. Thus the printed inequality has a signature/induced-metric convention built into it; the general formula above makes that choice explicit rather than losing the sign. The metric in the separate final question explicitly uses the mostly-plus signature.
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