Suppose I give you the following line element:
$$ ds^{2} = -(1+2\phi)dt^{2}+(1-2\phi)(dx^{2}+dy^{2}+dz^{2}) \tag{1}$$
Without saying anything you might think that this is just another line element like a line element in spherical coordinates; you might think that I just perform another coordinate transformation to a system who are endowned with a particular chart in the manifold $(M,g)$.
But, If I say to you that $(1)$ line element describes the Newtonian Gravity, and moreover, that the symbols $\phi$ are precisely the Newtonian potential of gravity, then you might ask: "why?" I would give you the following explanation:
The motion of a Freely Falling particle in a curved spacetime is then given by the following calculation of the absolute or intrinsic derivative: $$p^{b}\nabla_{b}p^{a} = p^{b}(\partial_{b}p^{a} + \Gamma ^{a}_{cb}p^{c})$$ If you consider the line element of $(1)$ and non-relativistic speeds then the equation of motion is reduced to (spatial components): $$p^{b}\nabla_{b}p^{a} = m\frac{d}{d\tau}p^{i} + \Gamma ^{i}_{00}(p^{0})^{2}$$ Which, after calculations is reduced to: $$\frac{dp^{i}}{d\tau} = -m \delta^{ij}\partial _{j}\phi = -m\partial _{i}\phi$$ Now, considering that we are in a regime of low velocities, then the Newtonian Equation is valid: $$F^{i} = -m(\nabla \phi_{g})^{i}$$ Then, since is also valid, by special relativity, that: $$ F^{i} = \frac{dp^{i}}{d\tau}$$ Then, $$-m(\nabla \phi_{g})^{i} = \frac{dp^{i}}{d\tau} = -m\partial _{i}\phi$$ Implies that $$\phi = \phi_{g}$$ And the line element describes the Newtonian gravity, indeed.
Now, given a reasonable explanation that $\phi = \phi_{g}$, you could say,also, that by equivalence principle $(1)$ also describes, in a local region, a relativistic form of gravity.
But here comes my doubt, I know that the geodesic deviation gives you the Riemann tensor, and for the metric $(1)$ you can conclude that the general quantity that encodes the notion of "gravity as curvature" is highly suggested by the Ricci tensor ,because the following expression gives you a way to talk about tidal effects:
$$a^{\mu}=\frac{D^{2} \delta x^{\mu}}{D\tau^{2}} = R^{\mu}_{\nu \gamma \beta} u^{\nu}u^{\gamma} \delta x^{\beta} \implies a^{i} = -c^{2}R^{i}_{0j0}\delta x^{j} $$
Where $ a^{\mu}=\frac{D^{2} \delta x^{\mu}}{D\tau^{2}}$ is called acceleration between geodesics, given by geodesic deviation.
And now, the doubt: Why tidal effects are non-local? Since we can describe tidal effects with the line element of $(1)$ by
$$a^{i} = -c^{2}R^{i}_{0j0}\delta x^{j} = -\frac{1}{c^{2}}\partial^{k} \partial_{k}\phi_{g}$$