Overview
The precession arises from the curvature of spacetime, not from a special relativistic length contraction. The first hint that the precession cannot be from special relativity (SR) is that in SR length contraction is parallel to the direction of motion but in this case, the separation used in Newtonian gravity is perpendicular to the direction of motion. In actuality, the precession arises from the curvature of space in the radial direction - below this is proved.
Proof
For a Schwarzchild spacetime the metric is:
$$g_{\mu\nu}=\operatorname{diagonal}\left(\begin{matrix}\left(1-\frac{2\mu}{r}\right)c^2,&-\left(1-\frac{2\mu}{r}c^2\right)^{-1},&-r^2,&-r^2sin^2\left(\theta\right)\end{matrix}\right)$$
where $\mu=\frac{GM}{c^2}$
working in the coordinates $x^\mu=\left(t,r,\theta,\phi\right)$
Due to the spherical symmetry, we need only consider orbits of $\theta=\frac{\pi}{2}$. Thus, the metric reduces to only a function of $r$.
Geodesics with affine parameterization $\lambda$ satisfy the Lagrangian:
$$L=g_{\mu\nu}\dot x^\mu\dot x^\nu$$
Thus, using the Euler-Lagrange equations we find:
$$\begin{align}\frac{\partial L}{\partial x^\rho}&=\frac{\text{d}}{\text{d}\lambda}\frac{\partial L}{\partial\dot x^\rho}\\\implies\frac{\partial g_{\mu\nu}}{\partial x^\rho}\dot x^\mu\dot x^\nu&=\frac{\text{d}}{\text{d}\lambda}\left[g_{\mu\nu}\left(\delta^\mu_\rho\dot x^\nu+\delta^\nu_\rho\dot x^\mu\right)\right]\end{align}$$
As the metric is diagonal and only depends on $r$: then for $\rho\ne1$:
$$\begin{align}0&=\frac{\text{d}}{\text{d}\lambda}\left[g_{\rho\rho}\dot x^\rho\right]\\\implies g_{\rho\rho}\dot x^\rho&=p_\rho\end{align}$$
where $p_0\equiv k,p_2=0,p_3\equiv h$ are constants and do not form a tensor
$$L=\frac{k}{g_{00}}+g_{11}\dot r^2+\frac{p_3}{g_{33}}$$
Using the explicit form for $\frac{p_3}{g_{33}}=\frac{h}{r^2}$ then $\dot r=-h\frac{\text{d}u}{\text{d}{\phi}}$ where $u=\frac{1}{r}$ and so:
$$\begin{align}L&=\frac{k}{g_{00}}+h^2g_{11}\left(\frac{\text{d}u}{\text{d}\phi}\right)^2+hu^2\\\implies\frac{L}{h^2g_{11}}&=\frac{k}{h^2g_{00}g_{11}}+\left(\frac{\text{d}u}{\text{d}\phi}\right)^2+\frac{u^2}{hg_{11}}\end{align}$$
Differentiating with respect to $\phi$ gives:
$$\begin{align}\frac{L}{h^2}\frac{\text{d}}{\text{d}\phi}\frac{1}{g_{11}}&=\frac{k}{h^2}\frac{\text{d}}{\text{d}\phi}\frac{1}{g_{00}g_{11}}+2\frac{\text{d}u}{\text{d}\phi}\frac{\text{d}^2u}{\text{d}\phi^2}+\frac{2u}{hg_{11}}\frac{\text{d}u}{\text{d}\phi}+\frac{u^2}{h}\frac{\text{d}}{\text{d}\phi}\frac{1}{g_{11}}\\\implies\frac{L}{h^2}\frac{\text{d}}{\text{d}u}\frac{1}{g_{11}}&=\frac{k}{h^2}\frac{\text{d}}{\text{d}u}\frac{1}{g_{00}g_{11}}+2\frac{\text{d}^2u}{\text{d}\phi^2}+\frac{2u}{hg_{11}}+\frac{u^2}{h}\frac{\text{d}}{\text{d}u}\frac{1}{g_{11}}\end{align}$$
$$\implies\frac{\text{d}^2u}{\text{d}\phi^2}+\underbrace{\frac{2u}{hg_{11}}+\frac{u^2}{h}\frac{\text{d}}{\text{d}u}\frac{1}{g_{11}}}_{=u-3\mu u^2}=\underbrace{\frac{L}{2h^2}\frac{\text{d}}{\text{d}u}\frac{1}{g_{11}}}_{=\frac{GM}{h^2}}-\underbrace{\frac{k}{2h^2}\frac{\text{d}}{\text{d}u}\frac{1}{g_{00}g_{11}}}_{=0}\tag{1}$$
Thus we get the differential equation:
$$\frac{\text{d}^2u}{\text{d}\phi^2}+u-3\mu u^2=\frac{GM}{h^2}\tag{2}$$
The Newtonian equation is:
$$\frac{\text{d}^2u}{\text{d}\phi^2}+u=\frac{GM}{h^2}$$
And so we can see the additional term that leads to precession is $-3\mu u^2$ which we can see, from inspecting equation (1), arises non-zero $\mu$ in the $g_{11}$ component of the metric. Additionally, note how the time component of the metric is such that it cancels with the radial component of the metric so there are no additional terms in the equation; thus, in a sense, the time component contributes to cancelling any additional effects but not to the precession.
The angular coordinates have the same metric components as Euclidean space due to isotropy and so would not be expected to affect the shape equation.
Modified Force
Now let's consider from a classical perspective what force would give rise to this orbit. Starting from conservation of energy:
$$\begin{align}E&=\frac{1}{2}m\left(\dot r^2+r^2\dot\phi^2\right)+V\left(r\right)\\\implies E&=\frac{1}{2}mh^2\left(\left(\frac{\text{d}u}{\text{d}\phi}\right)^2+u^2\right)+V\left(r\right)\end{align}$$
Using the same definitions of $u$ and $h$ as above. Now differentiating with respect to $\phi$ gives:
$$\frac{\text{d}^2u}{\text{d}\phi^2}+u=-\frac{1}{mh^2}\frac{\text{d}V}{\text{d}u}$$
Now as $F=-\frac{\text{d}V}{\text{d}r}$ then we get:
$$\frac{\text{d}^2u}{\text{d}\phi^2}+u=-\frac{F}{mh^2u^2}$$
Comparing this to equation (2) we can see to get the precession predicted by general relativity one would need the force law to be:
$$F=-\frac{GMm}{r^2}\left(1+\frac{3h^2}{c^2r^2}\right)$$
