How do I use the Schwarzschild metric to calculate space curvature and time curvature seperately? I want to understand the math behind the idea that around Earth time dilation accounts for 99.99% of gravity, while around a black hole it only accounts for 50% of gravity while space curvature accounts for the other 50%.  I think it can be shown with the first two terms of the Schwarzschild metric, but I can't figure out how.  Can you give some guidance on how I use the SM to show this discrepancy?  Or to put it in a different way; why — using the SM — does space curvature grow at such a rapid rate while time dilation does not when near a black hole?
 A: I quite dislike this particular turn of phrase - there is no meaningful sense in which curvature can be split into temporal and spatial parts.  We can talk about pure spatial curvature in the sense that we can foliate a $d$-dimensional spacetime into spacelike leaves $\Sigma_t$, each of which constitutes a $(d-1)$-dimensional Riemannian manfold. Each leaf inherits a Riemannian metric $\gamma$ from the pseudo-Riemannian metric $g$ of the full spacetime, and so we can talk about its inherited Riemann/Ricci tensors.
However, this is not really an intrinsic property of the full spacetime - a different foliation will result in different leaves with different curvatures.  More to the point, even in those situations where we can write the full spacetime as $\mathcal M = \mathbb R \times \Sigma$, a 1-dimensional manifold has no intrinsic curvature and therefore it doesn't make sense to talk about the "curvature of time."
What one presumably means when talking about this is the following.  If we consider the radial geodesic equation in Schwarzschild coordinates and assume only radial motion, we obtain the following equation (the dot denotes differentiation with respect to the proper time):
$$\ddot r + \Gamma^r_{\beta \nu} \dot x^\beta \dot x^\nu = 0 $$
$$\implies \ddot r =  -\Gamma^r_{tt} (c\dot t) ^2 - \Gamma^r_{r r} \dot r^2  - 2 \Gamma^r_{t r} (c\dot t) \dot r$$
Recall the formula for the Christoffel symbols:
$$\Gamma ^\mu_{\beta\nu} = \frac{1}{2} g^{\mu\rho} \big( \partial_\beta g_{\rho \nu} + \partial_\nu g_{\beta \rho} - \partial_\rho g_{\beta \nu}\big)$$
$$\implies \matrix{\Gamma^r_{tt} = \frac{1}{2} g^{rr} \big(\partial_t g_{rr} + \partial_t g_{rr} - \partial_r g_{tt}\big) = -\frac{1}{2}\big(1+\frac{2\Phi}{c^2}\big) \big(2\frac{\phi'}{c^2}\big)\\
\Gamma^r_{rr} = \frac{1}{2} g^{rr}\big(\partial_r g_{rr} + \partial_r g_{rr} - \partial_r g_{rr}\big)= \frac{1}{2}\big(1+\frac{2\Phi}{c^2}\big) \frac{-2\Phi'/c^2}{\big(1+\frac{2\Phi}{c^2}\big)^2}\\
\Gamma^r_{tr} = \frac{1}{2}g^{rr}\big(\partial_t g_{rr} +\partial_r g_{tr} - \partial_r g_{tr}\big) = 0 }$$
where $\Phi(r) = -GM/r$ is the Newtonian gravitational potential. After some simplification, the radial geodesic equation becomes
$$\ddot r = -\left(1+\frac{2\Phi}{c^2}\right)\frac{\Phi'}{c^2}\left(c^2\dot t^2 + \frac{\dot r^2}{\left(1+\frac{2\Phi}{c^2}\right)}\right)$$
$$= -\Phi'\left(1+\frac{2\Phi}{c^2}\right)\left(\dot t^2  + \frac{\dot r^2/c^2}{1+\frac{2\Phi}{c^2}}\right)$$
From the line element,
$$\mathrm d\tau^2 = \left(1+\frac{2\Phi}{c^2}\right) \mathrm dt^2 - \frac{1}{\left(1+\frac{2\Phi}{c^2}\right)} \mathrm dr^2$$
$$\implies \dot t^2 = \frac{1 + \frac{\dot r^2}{(1+2\Phi/c^2)}}{1+\frac{2\Phi}{c^2}}$$
and so finally
$$\ddot r = -\Phi'\left( 1 + \frac{2\dot r^2/c^2}{1+\frac{2\Phi}{c^2}}\right)$$
I assume that the author of your statement is referring to the relative magnitudes of the terms in this expression.
