Consider a 2d world. There are basically three useful cases to consider: Lorentzian (-,+) signature, Galilean (0, +) signature, and Euclidean (+, +) signature. [Yes, perhaps there is a (0,0) signature, but this is especially boring.]
Each of these spaces has a "rotation-like" operation, generated by exponentials of various quantities that are not real numbers but are geometrically significant to the space.
In Euclidean space, that quantity is something like $i$ in that it squares to $-1$ and thus, the rotation-like operation is rotation and the exponential is trigonometric in nature.
In Galilean space, that quantity is instead some $\epsilon$ such that $\epsilon^2 = 0$, and the exponential power series degenerates to $\exp(\epsilon) = 1 + \epsilon$. This generates the simple velocity-addition law of Galilean relativity.
In Lorentzian space(time), the fundamental quantity is some $j$ such that $j^2 =1 $. This is how the exponential generates hyperbolic functions; the rotation-like operation is Lorentz boosting.
What is unique to Lorentzian space is the presence of non-zero vectors that are unaffected by rotation-like operations in planes they occupy. Every vector in Euclidean space is rotated by some angle as long as it lies in the rotation plane. The same goes for Galilean space. But Lorentzian space has the lightlike or null vectors, which go along light cones, which are totally unaffected by boosts.
This is just part of what makes Lorentzian space(time) an attractive model for the physical world; it reduces the constancy of the speed of light to a purely geometric phenomenon. We need not ask ourselves why some object with a given speed should be perceived as having the same speed regardless of our orientation (reference frame). Choosing Lorentzian space(time) as our model makes the existence of such directions of travel patently obvious. The question then becomes why light should travel along those directions, which opens up new avenues of exploration for physics.
You should stop thinking of the metric tensor as a matrix itself. It is not. Matrices are only a device for facilitating computations, and you don't have a natural choice of basis for the tangent spaces to form a matrix for a metric given in an arbitrary manifold. In other words, to form a matrix, one needs to choose a coordinate system for the manifold.
There is nothing wrong in saying that the metric tensor in $\Bbb H^d_K$ is just the pull-back of the standard Lorentz metric of $\Bbb R^{d+1}_1$ via the inclusion $\Bbb H^d_K\hookrightarrow \Bbb R^{d+1}_1$, and this just happens to be Riemannian metric. I don't see why one would insist on using a $d+1$ order matrix to represent a metric in a manifold of order $d$. The matrix would be the same, ${\rm diag}(-1,1,\ldots,1)$, the only difference being that now this matrix accepts fewer inputs. In other words, it's like using a matrix representation of a linear map to compute it's restriction to a subspace but promising you won't evaluate it in vectors outside your subspace.
As far as understanding the geometry of the submanifold, there are much more efficient ways, bypassing this awkward approach with matrices. To compute the sectional curvature of $\Bbb H^d_K$, though, is by using the following consequence of the Gauss formula (which you can see in any Riemannian geometry book, and remains valid in the pseudo-Riemannian case): $$K(X,Y) =\overline{K}(X,Y) +\frac{\langle \alpha(X,X),\alpha(Y,Y)\rangle -\langle\alpha(X,Y),\alpha(X,Y)\rangle}{\langle X,X\rangle\langle Y,Y\rangle},$$where $K$ and $\overline{K}$ denote the sectional curvatures of the submanifold and ambient manifold, and $\alpha$ denotes the second fundamental form, satisfying $\overline{\nabla}_XY=\nabla_XY+\alpha(X,Y)$ for all vector fields tangent to the submanifold. Here $\nabla$ and $\overline{\nabla}$ denote the Levi-Civita connections of the submanifold and of the ambient manifold.
Now, note that $\xi(p) = \sqrt{-K}p$ is a unit normal timelike vector to $\Bbb H^d_K$. So $$\alpha(X,Y)=-\langle \alpha(X,Y),\xi\rangle\ \xi = -\langle A_{\xi}(X),Y\rangle \xi = \langle \overline{\nabla}_X\xi,Y\rangle \xi = \sqrt{-K}\langle X,Y\rangle \xi.$$So if $(X,Y)$ is an orthonormal basis for any $2$-plane tangent to $\Bbb H^d_K$, we have $$K(X,Y) = \langle \sqrt{-K}\xi,\sqrt{-K}\xi\rangle = -K\langle \xi,\xi\rangle = K.$$
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Here you need to think of the pair (surface, metric) as the geometric object, not only of the surface. The upshot here is that the pairs (hyperboloid, euclidean metric) and (hyperboloid, Minkowski metric) are distinct geometric objects. The way to measure lengths and areas is not the same in these two "worlds". The first one has non-constant positive curvature, and the second one has constant negative curvature $-1$.
The reason is that the hyperboloid plays the role of a sphere with respect to the Minkowski metric, since it can be written as the solution set of $\langle p,p\rangle_L=-1$. Compare that with the sphere equation $\langle p,p \rangle_E=1$. If by now you're wondering whether the one-sheeted hyperboloid defined by $\langle p,p\rangle_L=1$ has something special about it, I'll tell you it does: it has constant curvature $1$ when equipped with the Minkowski metric (the one who correctly expresses it as a sphere).
In fact, here's an exercise for you to understand what happens: define $$g((x_1,y_1,z_1),(x_2,y_2,z_2))= 3x_1x_2+5y_1y_2+2z_1z_2$$and show that $M=\{(x,y,z)\in \Bbb R^3\mid 3x^2+5y^2+2z^2=1\}$ has constant curvature when equipped with $g$.
Solution for suggested exercise. Note that $M = \{ p \in \Bbb R^3\mid g(p,p)=1\}$. If $F(p) = g(p,p)$, then $M = F^{-1}(1)$, meaning that the $g$-gradient of $F$ is always $g$-normal to $M$. Since $${\rm d}F_p(v) = 2g(p.v) = g(2p,v),$$we get ${\rm grad}_gF(p)=2p$, and so $N(p)=p$ is an $g$-unit $g$-normal vector along $M$, just like for spheres in Euclidean space. Indeed, $M$ is a $g$-sphere. Note that the curvature of the Levi-Civita connection of $g$ in $\Bbb R^3$ is zero (indeed, all the Christoffel symbols w.r.t. the usual rectangular coordinates are zero -- this actually says that the Levi-Civita connection is the usual one). So, Gauss' Formula says that $$K(v,w) = \frac{g(I\hspace{-.1cm}I(v,v))g(I\hspace{-.1cm}I(w,w)) - g(I\hspace{-.1cm}I(v,w),I\hspace{-.1cm}I(v,w))}{g(v,v)g(v,w)-g(v,w)^2}.$$Just in the same way we do it for the sphere, you can check (or see e.g. page 101 in Chapter 4 in O'Neill's Semi-Riemannian Geometry with Applications to Relativity) that $I\hspace{-.1cm}I_p(v,w) = -g(v,w)N(p)$. Taking $(v,w)$ to be an orthonormal basis for the tangent plane $T_pM$, plugging in above gives $K(p)=1$ as expected.