Newton's Law of Universal Gravitation tells us that the potential energy of object in a gravitational field is $$U ~=~ -\frac{GMm}{r}.\tag{1}$$
The experimentally verified near-Earth gravitational potential is $$U ~=~ mgh.\tag{2}$$
The near-Earth potential should be an approximation for the general potential energy when $r\approx r_{\text{Earth}}$, but the problem I'm having is that they scale differently with distance. $(1)$ scales as $\frac 1r$. So the greater the distance from the Earth, the less potential energy an object should have. But $(2)$ scales proportionally to distance. So the greater the distance from the Earth, the more potential energy an object should have.
How is this reconcilable?
Answer
Your equation (2) is the change in potential energy when the object moves vertically by a distance $h$ i.e. when the object moves from $r$ to $r+h$. Let's use equation (1) to calculate this:
$$ \Delta U = GMm\left(\frac{1}{r}-\frac{1}{r+h}\right) $$
Subtracting the two fractions inside the bracket gives:
$$\begin{align} \Delta U &= GMm\left(\frac{r+h}{r(r+h)}\frac{r}{r(r+h)}\right) \\ &= GMm\frac{h}{r(r+h)} \end{align}$$
Since $h \ll r$ that means $r+h\approx r$ and our equation becomes:
$$\begin{align} \Delta U &\approx GMm\frac{h}{r^2} \\ &\approx \frac{GM}{r^2}mh \\ &\approx gmh \end{align}$$
Footnote:
I've just noticed that in your comment to Itachí's answer you ask if you can use a Taylor series. You can use a binomial expansion to make the aproximation more obvious. You rewrite:
$$ \Delta U = GMm\frac{h}{r(r+h)} $$
as:
$$ \Delta U = \frac{GM}{r^2}mh\left(1+\frac{h}{r}\right)^{-1} $$
then a binomial expansion gives:
$$ \Delta U = \frac{GM}{r^2}mh\left(1-\frac{h}{r} + O\left(\frac{h}{r}\right)^2 \right) $$
And as before since $h \ll r$ the term in the brackets is approximately one and we once again get:
$$ \Delta U = \frac{GM}{r^2}mh $$
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