= Solution
Assume the level set is regular and non-null. The latter hypothesis is necessary: a <null hypersurface> has a degenerate <induced metric> and hence no ordinary intrinsic <Levi-Civita connection> determined by that metric. Normalize the <normal vector> and define the <non-null hypersurface projection> by
$$
n_a=\frac{\partial_af}{\sqrt{|g^{bc}\partial_bf\partial_cf|}},\qquad \varepsilon=n_an^a=\pm1,\qquad P^a{}_b=\delta^a{}_b-\varepsilon n^an_b,\qquad h_{ab}=g_{ab}-\varepsilon n_an_b.
$$
Here $h$ denotes the <induced metric> restricted to <tangent vectors>, while its ambient extension annihilates $n$. The intrinsic <covariant derivative> is $D_XY=P\nabla_XY$ for tangent <vector fields>. Projection of the ambient <torsion-free connection> and <metric compatibility> shows that $D$ is the <Levi-Civita connection> of $h$. Define the <extrinsic curvature> by $K(X,Y)=g(\nabla_Xn,Y)$, or $K_{ab}=P^c{}_aP^d{}_b\nabla_cn_d$. It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric <Hessian>. This sign convention makes the scalar <second fundamental form> $g(n,\nabla_XY)=-K(X,Y)$.
Let $S$ be the tangent linear map characterized by $h(SX,Y)=K(X,Y)$. Differentiating $g(n,n)=\varepsilon$ and $g(n,Y)=0$ gives the <Gauss formula>
$$
\nabla_Xn=SX,\qquad \nabla_XY=D_XY-\varepsilon K(X,Y)n.
$$
Fix the <Riemann curvature tensor> convention
$$
R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z,\qquad R_{abcd}=g(e_a,R(e_c,e_d)e_b).
$$
Applying the <Gauss formula> twice gives
$$
P\nabla_X\nabla_YZ=D_XD_YZ-\varepsilon K(Y,Z)SX.
$$
Subtracting the expression with $X,Y$ interchanged, and then subtracting the bracket derivative, proves
$$
\mathcal R(X,Y)Z=PR(X,Y)Z+\varepsilon\bigl(K(Y,Z)SX-K(X,Z)SY\bigr).
$$
Thus the <Gauss–Codazzi equations for a non-null hypersurface> give the intrinsic <Riemann curvature tensor> as
$$
\boxed{\mathcal R_{abcd}=P^e{}_aP^f{}_bP^g{}_cP^h{}_dR_{efgh}+\varepsilon(K_{ac}K_{bd}-K_{ad}K_{bc}).}
$$
All free indices here are tangent to the <hypersurface>. The normal component of the same calculation is
$$
g(n,\nabla_X\nabla_YZ)=-K(X,D_YZ)-X\bigl(K(Y,Z)\bigr).
$$
After antisymmetrizing and using $D_XY-D_YX=[X,Y]$, this yields the curved-ambient <Codazzi equation>
$$
\boxed{g(n,R(X,Y)Z)=-(D_XK)(Y,Z)+(D_YK)(X,Z).}
$$
The <extrinsic curvature> measures how tangent planes turn in the ambient manifold; the <Gauss equation> says that this turning contributes to intrinsic curvature even when the ambient <Riemann curvature tensor> is zero.
For the quadric in <Minkowski spacetime>, write its position vector as $X^a$. On the <hypersurface>, $g(X,X)=1$ and differentiating this identity in a tangent direction gives $g(X,Y)=0$. Hence $n=X$ is a <unit normal> and a <spacelike vector>, $\varepsilon=1$, and $\nabla_Yn=Y$ in flat <Cartesian coordinates>. Therefore $K=h$. The ambient <Riemann curvature tensor> vanishes, so
$$
\boxed{\mathcal R_{abcd}=h_{ac}h_{bd}-h_{ad}h_{bc},\qquad \mathcal R_{ab}=3h_{ab},\qquad \mathcal R=12>0.}
$$
This is unit-radius <de Sitter spacetime>. Reversing the normal changes $K$ to $-K$ but does not change this curvature. An opposite overall convention for the <Riemann curvature tensor> would reverse both curvature signs below.
To retain a four-dimensional <Lorentzian metric> while reversing the curvature, take a flat ambient metric with two negative directions and the quadric
$$
-dt^2-dw^2+dx^2+dy^2+dz^2,\qquad -t^2-w^2+x^2+y^2+z^2=-1.
$$
Now the position vector is a <unit normal> with negative squared norm, so $\varepsilon=-1$ while $K=h$ remains true. The <Gauss equation> gives
$$
\boxed{\mathcal R_{abcd}=-(h_{ac}h_{bd}-h_{ad}h_{bc}),\qquad \mathcal R=-12.}
$$
This is unit-radius <Anti-de Sitter spacetime>. If a positive-definite <induced metric> is acceptable, an even simpler alternative is to keep the original one-time ambient <Minkowski metric> and change the quadric level to $g(X,X)=-1$. Each sheet then gives four-dimensional <hyperbolic space> with the same negative curvature, but its induced signature differs from that of <de Sitter spacetime>.
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