What explains the descrepancy between Asaph Hall's equation for Mercury's precession and GR? In regard to What is the weight equation through general relativity?, the answer is:
$$F=ma=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}.$$
This source provides a different equation:
$$f \approx - \frac{G M_0}{r^2} - \frac{3 G M_0 h^2}{c^2 r^4}.$$
And these two results go in the same direction of increasing the acceleration over Newton's equations. I'm not concerned with reconciling the differences between those two equations.
But Asaph Hall found (source):

Hall noted that he could account for Mercury's precession if the law of gravity, instead of falling off as $\frac{1}{r^2}$, actually falls of as $\frac{1}{r^n}$ where the exponent $n$ is $2.00000016$.

This would result in reduced acceleration compared to Newton's equations, and hence goes in the opposite direction suggested by GR.
It is a curious thing that either an increasing or a decrease in the acceleration could explain the motions of Mercury.  Now, I know that Asaph Hall's finding was not found to be correct as a theory for gravity, and that GR has been.  But as far as I can tell, his math wasn't wrong for the specific case of Mercury.
What explains how both GR and Asaph could explain Mercury's motion?
 A: Hall changed the proportionality for the force, but you need to know the constant of proportionality in order to say whether the gravitational force was stronger or weaker.
If the gravitational constant is fixed by (e.g.) requiring the Earth to have an orbital period of 1 year at 1 au, then having $n>2$ would require an increase in the constant of proportionality in the force equation.
If this constant of proportionality is then applied to an object at smaller orbital radii than the Earth then the gravitational force would be larger than the Newtonian value.
If we have a look at Hall (1894), where the original suggestion was made, you will see on p.50 that a new parameter is defined
$$ \mu = \mu' r^\Delta\ ,$$
where $\Delta = 0.00000016$ represents the departure of the force scaling from the inverse square law and $\mu$ is the gravitational constant for the Sun and Mercury.
In that case, the acceleration at the position of Mercury is given by
$$ a = \frac{\mu}{r^{2+\Delta}} = \frac{\mu'}{r^{2}}\ . $$
However, I don't think one cares what the strength of the gravitational field is. What matters in terms of whether and in what direction the precession is, is how the force varies with $r$. A general expression for the factor by which the apsidal angle$^1$ differs from $\pi$ is (see here)
$$ \psi = \pi\left( 3 + r \frac{F'(r)}{F(r)}\right)\ ,$$
which does not depend on the magnitude of $F(r)$.
$^1$ The angle through which the radius vector rotates between aphelion and perihelion - which would be $\pi$ for a Keplerian orbit.
