Gravity must be understood as a curvature of spacetime rather than space itself because the 1915 general theory of relativity, Einstein's new theory of gravity, is an extension of the 1905 special theory of relativity and the special theory of relativity introduces an inseparable connection between the space and the time and forces us to talk about them in a unified – talk about spacetime.
Space and time have to mix according to special relativity because the theory starts from two postulates, including the absolute constancy of the speed of light in the vacuum, and if space and time were separated, such a constancy would be incompatible with the other postulate, the identical form of the physical laws as seen by an arbitrary inertial observer. It makes no sense to discuss a better, post-Newtonian theory of gravity without taking special relativity into account; the general theory of relativity with its insights about the spacetime curvature is a result of the reconciliation of Newton's gravity and special relativity.
In fact, when one studies how Newton's approximate (inverse square) laws of gravity emerge from general relativity, it turns out that the "curvature of time", and not so much "curvature of space", as a function of space plays the decisive role in determining the gravitational fields at each point. Technically speaking, the rate of time at a given point is determined by $g_{00}$ which is approximately a linear function of the gravitational potential $\Phi$ known from Newton's theory.
Conceptual ideas may precede the mathematical formulation of some principles but one usually can't get too far if he avoids mathematics. Well over 99% of important insights in modern physics depend on mathematical equations and structures that may be at most translated to "awkward and confusing" words.
As suggested by ANDREW, I will write up the answer to the question here myself.
Given the hypersurface $\Sigma = \{ t = f(r) \}$, the induced metric is obtained by replacing $dt = df$ within the expression for the Schwarzschild metric $g$. The induced metric turns out to be
$$h \enspace = \enspace \Big[ \tfrac{1}{a(r)} - a(r) \, \big( \tfrac{\partial f}{\partial r} \big)^2 \Big] \, dr^2 + r^2 \, d\Omega^2$$
Following ANDREW's and JAVIER's proposal, I now identify the induced metric with some flat metric I already know. Since I am using spherical coordinates, the Euclidean metric in spherical coordinates volunteers for such an approach. In order for the induced metric to coincide with the Euclidean metric in spherical coordinates, the property
$$ \Big[ \tfrac{1}{a(r)} - a(r) \, \big( \tfrac{\partial f}{\partial r} \big)^2 \Big] \enspace = \enspace 1$$
must hold. This leads to
$$ \frac{\partial f}{\partial r} \enspace = \enspace \sqrt{\frac{1-a(r)}{a^2(r)}} \enspace = \enspace \sqrt{\frac{\tfrac{2m}{r}}{\big(1-\tfrac{2m}{r} \big)^2}} \quad ,$$
where $a(r) = 1 - \tfrac{2m}{r}$. The solution for this integral is
$$ f \enspace = \enspace 2m \cdot \bigg( 2\sqrt{\tfrac{r}{2m}} + \ln \Big( \big| 1 - \sqrt{\tfrac{r}{2m}} \big| \Big) - \ln \Big( 1 + \sqrt{\tfrac{r}{2m}} \Big) \bigg)$$
up to some constant. The hypersurface is now the set of points with coordinates $(t,r,\theta,\phi) = (f(r), r, \theta, \phi)$, where $f$ is the above function.
Best Answer
In general, it is the Riemann tensor that encodes curvature, not the metric. Although it is quite difficult to see why Riemann tensor describes curvature directly from its definition, due to its abstractness, it is fairly easy to see it geometrically from the equivalent notion of sectional curvature (https://en.wikipedia.org/wiki/Sectional_curvature).
Fortunately, in theories with Levi-Civita connection (torsionless and metric compatible), like General Relativity, the Christoffel symbols are given in terms of the metric (and its derivatives of course) and, in turn, the Riemann tensor is given as a function of the metric. Only in this case that Riemann tensor is a function of the metric.