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.
Habouz asked: "What's the Schwarzschild metric in terms of a falling observer's coordinates?"
The transformation rule can be found here and here:
$${\rm dt = d\tau+dr \ v} / \hat g_{\rm tt} \ , \ \ {\rm v = -c \ \sqrt{r_s/r}} \ , \ \ 1/\gamma = \rm \sqrt{1-v^2/c^2}$$
with that you transform the old coordiantes $\rm \hat x$ to the new ones $\rm \bar x$:
$$\bar g_{\mu \nu} = \sum_{\sigma, \kappa} \ \hat g_{\sigma \kappa} \ \rm \frac{\partial \hat x^{\sigma}}{\partial \bar x^{\mu}} \ \frac{\partial \hat x^{\kappa}}{\partial \bar x^{\nu}}$$
so the metric in regular Schwarzschild/Droste coordinates (with time $\rm \hat x^0 = t$) where the local observers are stationary with respect to the black hole:
$$\hat g_{\mu \nu} = \left(
\begin{array}{cccc}
\rm c^2/\gamma^2 & 0 & 0 & 0 \\
0 & \rm -\gamma^2 & 0 & 0 \\
0 & 0 & \rm -r^2 & 0 \\
0 & 0 & 0 & \rm -r^2 \sin ^2 \theta \\
\end{array}
\right)$$
transforms to Gullstrand/Painlevé coordinates (with time $\rm \bar x^0 = \tau$) where the local observers are free falling raindrops with the negative escape velocity $\rm v$:
$$\bar g_{\mu \nu} = \left(
\begin{array}{cccc}
\rm c^2/\gamma^2 & \rm v & 0 & 0 \\
\rm v & \rm -1 & 0 & 0 \\
0 & 0 & \rm -r^2 & 0 \\
0 & 0 & 0 & \rm -r^2 \sin ^2 \theta \\
\end{array}
\right)$$
where the covariant spatial components are euclidean:
$$\bar g_{\rm i j} = \left(
\begin{array}{cccc}
\rm -1 & 0 & 0 \\
0 & \rm -r^2 & 0 \\
0 & 0 & \rm -r^2 \sin ^2 \theta \\
\end{array}
\right)$$
which is just flat space in spherical $\{ \rm r, \ \theta, \ \phi \}$ coordinates:
Hamilton & Lisle wrote: "In the river model, space itself flows like a river through
a flat background, while objects move through the river according to the rules of special relativity"
Here are two such free falling raindrops in Gullstrand/Painlevé coordinates sending signals to each other when they are half way through the black hole (at $\rm r=r_s/2=1$, photons in green):
If you take ${\rm dt = du}\pm{\rm dr \ c} / \hat g_{\rm tt}$ or alternatively ${\rm dt = d{}_T\pm dr \ v^2/c} / \hat g_{\rm tt}$ instead (with $\rm {}_T=u \mp r$) you get the Schwarzschild metric in ingoing or outgoing Eddington/Finkelstein coordinates where radial photons have a constant coordinate velocity of $\rm dr/d{}_T=-c$, while in Gullstrand/Painlevé coordinates the free fallers have a coordinate velocity of $\rm dr/d\tau=v$.
In regular Schwarzschild/Droste coordinates on the other hand, at $\rm r=r_s$ you get $\rm dr/dt=0$ for all particles and photons since the bookkeeper's coordinate time $\rm t$ freezes at the horizon.
Best Answer
Neither the metric nor the curvature depend on the coordinates. They are tensors. They exist independent of coordinates and are invariant under any coordinate transformation. e.g. $\boldsymbol{\eta}$ is the metric tensor for Minkowski space no matter your choice of coordinates. However, you can project a tensor onto a particular coordinate basis to obtain its $\textit{coordinate-dependent}$ components.
In Cartesian coordinates, $\boldsymbol{\eta}$ has components $\eta_{\alpha\beta} = \text{diag}(-1, 1, 1, 1)$. In spherical coordinates, $\boldsymbol{\eta}$ has components $\eta_{\bar{\mu}\bar{\nu}} = \text{diag}(-1, 1, r^2, r^2\sin^2\theta)$. At the end of the day, these components still represent the same geometric object:
$$\begin{align} \boldsymbol{\eta} &= -1\tilde{dt}\otimes\tilde{dt}+1\tilde{dx}\otimes\tilde{dx}+1\tilde{dy}\otimes\tilde{dy}+1\tilde{dz}\otimes\tilde{dz} \;\;\;\;\text{(Cartesian)}\\ &= -1\tilde{dt}\otimes\tilde{dt}+1\tilde{dr}\otimes\tilde{dr}+r^2\tilde{d\theta}\otimes\tilde{d\theta}+r^2\sin^2\theta\tilde{d\phi}\otimes\tilde{d\phi} \;\;\;\;\text{(spherical)}\\ &=\eta_{\alpha'\beta'}\tilde{d}x^{\alpha'}\otimes\tilde{d}x^{\beta'} \;\;\;\;\text{(arbitrary coordinates)} \end{align}$$
Similarly the curvature tensor is the same in all coordinate systems. A straightforward calculation shows the curvature is the zero tensor in Minkowski space - no matter the coordinate system you choose, if any.
Nonetheless, we often subscribe to a particular coordinate system to help with calculations. Then the Riemann tensor would have coordinate-dependent components: $R_{\alpha\beta\mu\nu}$. To study the curvature without being fooled by coordinate artifacts, we can compute curvature invariants which are scalars and the same in every frame.
The Kretschmann scalar $K$ is a particular curvature invariant. In the Schwarzschild geometry,
$$K = R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta} = \frac{48M^2}{r^6}$$
where we've used Schwarzschild coordinates: $(t,r,\theta,\phi)$. So the curvature is finite at the event horizon ($r=2M$), and diverges as $r\rightarrow 0$.