To arrive at a non-relativistic limit, it is necessary first to divorce the connection between the covariant and contravariant metric in Relativity, treating them, instead, as two otherwise-independent objects that happen to be invariants of the same local symmetry group, that group being the Lorentz group for the metrics of space-times with a 3+1 signature.
In particular, the metric for 3+1 dimensional Minkowski space, endowed with coordinates $\left(x^1,x^2,x^3,x^4\right) = \left(x,y,z,t\right)$, may be included in the following family of metrics given by the following geometric objects:
$$g^{μν} {∂ \over ∂x^μ} {∂ \over ∂x^ν} = β ∇^2 - α \left({∂ \over ∂t}\right)^2,$$
$$g_{μν} dx^μ dx^ν = β dt^2 - α \left(dx^2 + dy^2 + dz^2\right),$$
which are then taken to be invariants, as well as the identity tensor:
$$δ^μ_ν dx^ν {∂ \over ∂x^μ} = dx {∂ \over ∂x} + dy {∂ \over ∂y} + dz {∂ \over ∂z} + dt {∂ \over ∂t}$$
where
$$∇^2 ≡ {∂ \over ∂x}^2 + {∂ \over ∂y}^2 + {∂ \over ∂z}^2$$
and where I am using the Einstein summation convention here and below. The contraction of these two is a multiple of the identity tensor:
$$g^{μρ}g_{ρν} = αβδ^μ_ν.$$
For the cases $\left(α,β\right) ≠ \left(0,0\right)$, the symmetry group which has both the metric and dual metric as invariants, and identity tensor, has dimension 6. When $αβ > 0$, it is the Lorentz group (that is: the homogeneous part of the Poincaré group), and the corresponding geometry is Minkowski geometry, with light speed given by $c = \sqrt{β/α}$. When $αβ < 0$, it SO(4), the homogeneous symmetry group for the geometry with 4+0 signature - 4-dimensional Euclidean space. Apart from the exception above, the cases for $αβ = 0$, are $α ≠ 0$ and $β = 0$, which corresponds to the $c → 0$ limit - the homogeneous part of the Carrollian group and its corresponding geometry - and $α = 0$ and $β ≠ 0$, the $c → ∞$ limit - the homogeneous part of the Galilei group and a geometry for the Newtonian paradigm. Both of these have degenerate signatures. Finally, although the invariants vanish in the $\left(α,β\right) = \left(0,0\right)$ case, the corresponding symmetry groups have a non-trivial limit as $\left(α,β\right) → \left(0,0\right)$: that is the homogeneous part of the Static group.
For the non-degenerate signatures, both the metric and the dual metric are rank 4 and non-singular and contract to a constant multiple of the identity tensor, so they can be adjusted so as to be inverses of one another.
For the degenerate signatures, they each have lower rank: rank 3 for metric and rank 1 for the dual metric in the $c → 0$ limit; and rank 1 for the metric and rank 3 for the dual metric in the $c → ∞$.
So, in the general case, a Newton-Cartan geometry has a metric and dual metric that are respectively ranks 1 and 3, and whose contraction is the 0 tensor.
Newton-Cartan geometry is formulated wrong. It has both the metric and its dual:
$$g^{μν} {∂ \over ∂x^μ} {∂ \over ∂x^ν} = ∇^2,$$
$$g_{μν} dx^μ dx^ν = dt^2,$$
where $\left(α,β\right)$ is normalized to $\left(0,1\right)$. You don't see that, so clearly, because the metric - by virtue of being only rank 1 - is factored into the square of a 1-form $dt$, which is then treated as a separate fundamental object in place of the metric; and the curved variants of the geometry have a metric and dual metric that reduce locally to these two.
Part of what makes this formulation wrong is that it only accounts for what might be called globally parabolic geometries - those in which the rank 1 metric has a global decomposition into the tensor product square of the 1-form, $dt$ (which is generalized to the 1-form $ψ$). A rank 1 metric need not decompose globally to the tensor product square of a one-form or the tensor product of two one-forms. Such cases could, then, be termed the globally non-parabolic geometries.
As a point of interest, just as the symmetry groups can all be treated as deformations of the Static group, with $α$ and $β$ being the 2 deformation parameters, so can the geometries. This requires going one dimension higher. The simplest way to arrive at this extension is to just simply treat the proper time - denoted here by $s$ - as a separate coordinate in itself. To make this unification consistent for $α = 0$ case, we should instead regard the time dilation $s - t$ as the extra coordinate, but adjust it downward by a factor of $α$, defining
$$u ≡ {{s - t} \over α},$$
i.e. by decomposing $s = t + αu$.
This has a consistent and non-trivial limit as $α → 0$, despite the fact that $s = t$ in that limit. To put it differently and more directly:
$$\mbox{There is a non-relativistic version of Time Dilation!}$$
as embodied by the $u$ coordinate. That extra coordinate can be added to the others to identify the coordinates $\left(x^1,x^2,x^3,x^4,x^5\right) = \left(x,y,z,t,u\right)$ of a 5-dimensional geometry which, in the non-relativistic case, is called the Bargmann geometry.
The corresponding geometric invariants can be extracted by substituting for $ds$ in for the proper time line element
$$βds^2 = βdt^2 - α\left(dx^2 + dy^2 + dz^2\right)$$
and dividing out by $α$ to obtain the following invariant condition
$$dx^2 + dy^2 + dz^2 + 2β dt du + αβ du^2 = 0,$$
with the corresponding dual metric being given by the invariant object
$$β ∇^2 + 2 {∂ \over ∂t}{∂ \over ∂u} - α \left({∂ \over ∂t}\right)^2.$$
Their contraction is now reduced by a factor of $α$ to a factor $β$ of the 5-D identity tensor
$$δ^μ_ν dx^ν {∂ \over ∂x^μ} = dx {∂ \over ∂x} + dy {∂ \over ∂y} + dz {∂ \over ∂z} + dt {∂ \over ∂t} + du {∂ \over ∂u}.$$
If we also require that $ds$ be an invariant, then it follows that $∂/∂u$ must also be an invariant.
The symmetry groups that leave the quadratic invariants (the metric and its dual), as well as the linear invariants and identity tensor fixed are the central extensions of the respective symmetry groups, though I'm not sure this fully characterizes the centrally extended Static group in the $\left(α,β\right) = \left(0,0\right)$ case.
In the non-relativistic case, this is called the Bargmann group, corresponding to the Bargmann geometry, and the central extension is non-trivial, as it is for the Static group. For the other cases, the central extensions are trivial and reduce to a direct product.
The Bargmann geometry provides a more consistent foundation for the non-relativistic paradigm and its generalization to curved geometries, than does Newton-Cartan; but it wasn't known at the time Cartan originally laid out the formulation for Newton-Cartan geometries, otherwise he probably would have used that, instead.
An example that illustrates the Schwarzschild solution directly as a deformation of the geometry for Newtonian gravity, where we restrict $β = 1$ and allow $α$ to vary from $α = 0$ to $α = 1/c^2$. Write the metric in the following form as a proper time metric:
$$ds^2 = (1 + 2αV) dt^2 - α\left(dx^2 + dy^2 + dz^2 - {2αV \over {1 + 2αV}} dr^2\right)$$
and then substituting $s = t + αu$ and dividing out by $α$ to obtain
$$dx^2 + dy^2 + dz^2 + 2 dt du + α du^2 - 2V dt^2 - {2αV \over {1 + 2αV}} dr^2 = 0,$$
where we continue to use use Cartesian coordinates for convenience, but also the radial coordinate $r ≡ \sqrt{\left(x^2 + y^2 + z^2\right)}$, where $r dr = x dx + y dy + z dz$. This is the metric for a gravitating body at $\left(x,y,z\right) = \left(0,0,0\right)$ with mass $M$ and gravitational potential $V = -GM/r$.
In this decomposition, the first three terms are those for Euclidean 3-dimensional geometry, the fourth is for Galilean Relativity, the fifth is the Special Relativity terms, whose inclusion converts Galilean Relativity to that of the Lorentz group, the sixth term is the Newtonian Gravity term, and the seventh and last term is the tiny correction - characteristic of General Relativity - that accounts for the warping of space, itself, such as what would be measured by the drift of a gyroscope taken on repeated orbits about the Earth.
In the limit as $α → 0$, this becomes
$$dx^2 + dy^2 + dz^2 + 2 dt du - 2V dt^2 = 0.$$
That is the line element for Newtonian Gravity, as represented in Bargmann geometry.
To find out what $u$ is, write down the corresponding light cone condition for the velocity of a test mass:
$$\dot{x}^2 + \dot{y}^2 + \dot{z}^2 + 2\dot{u} - 2V = 0,$$
or
$$\dot{u} = V - {v^2 \over 2}$$
where $𝐯 ≡ \left(\dot{x},\dot{y},\dot{z}\right)$ is the velocity of the body and $v^2 = \left|𝐯\right|^2$. Up to sign, $\dot{u}$ is the Lagrangian per unit mass for the test body, so $u$ is the (negative) action per unit mass for the body.
