Why does a curved surface result in a larger accumulation of charge, compared to a flat surface? Recently I attended an introductory electromagnetism course where it was stated that a curved conductor surface can sustain a larger charge density than a flat conductor surface. Conventionally this phenomenon is explained by asserting that the parallel repulsive electric force between the electrons is smaller than at a level surface. However this parallel repulsive force is seemingly always caused by two oppositely located electrons (see the image below); but what about the repulsive force between neighboring electrons? To me it seems logical that the repulsive force between neighboring electrons, on a curved surface, must be far greater than that of neighboring electrons on a flat surface. Simply because those electrons are closer and therefore the Coulomb force must be greater.
All in all I would believe that the electrons on the flat surface should actually be closer to one another, however this is not the case. The conventional explanation seems incomplete/missing a few details, or perhaps I'm missing some key points.

Image source https://courses.lumenlearning.com/physics/chapter/18-7-conductors-and-electric-fields-in-static-equilibrium
 A: 
Charges repel along straight line connecting them. Consider the component of repulsive force along the surface, which affects the spacing. Less curvature: net force of one charge on its neighbor is mainly along the surface so charges move further from each other. High curvature - net force is at a greater angle to the surface and charges can be closer to each other as smaller component of the force is directed along the surface.
You are thinking in terms of net force but only the parallel component is important here; electrons on a conductor sit at the surface so perpendicular component is irrelevant to their spacing.
A: This is a confusing topic, because two contradictory-sounding statements about the charge buildup on the surface of a conductor are both true.

*

*The surface charge density $\sigma$ is highest where the surface is most curved.

*The total charge $Q$ is still mostly located where the surface is least curved.

The first statement turns out to be the more important one for most practical purposes, since the surface charge density is directly related to the electric field immediately outside the conductor, $\vec{E}=\frac{\sigma}{\epsilon_{0}}\hat{n}$.  This means, in particular, that sparks tend to emerge from (or arc to) the more pointed parts of a conductor, since the large surface fields there make it easy to ionize the air, producing a conductive path for a charge discharge to follow.  (This is the basis for how lightning rods function, but providing a grounded, sharply-pointed conductor for the charged clouds to discharge to.)
The resolution of the apparent conflict between the two bulleted points above is that regions of a conducting surface where there is a high degree of curvature have relatively small total surface areas.  (If this is confusing, just think of a sphere.  The natural measure of curvature is the inverse of the radius of curvature, $k=1/R$.  The larger $k$ is, the smaller the surface area $4\pi/k^{2}$ is.) In the limiting case of a cusp, there is actually vanishing surface are at the point. The surface area where the curvature is small is greater, so if we consider the amount of charge $\Delta Q$ contained on some small patch of area $\Delta A$, there is a competition in $\Delta Q=\sigma(\Delta A)$ between $\sigma$, which increases with $k$, and $\Delta A$, which decreases with $k$.  Ultimately, in $\Delta Q$, the dependence of $\Delta A$ on $k$ wins out, and areas with greater curvature have less total charge (in spite of having a larger $\sigma$).
This can be seen in figures such as this, showing the field lines around a charged teardrop-shaped conductor.

For the teardrop conductor, the density of field lines (and thus $E$) is much greater near the cusp, but the total number of emerging field lines (and thus the total charge) is greater on the larger, smoother part of the teardrop.  (It is also useful to keep in mind that this figure shows only a two-dimensional cutaway; the imbalance in number of field lines is ever more evident if the full three-dimensional picture is shown.)  The concentration of charge that produces a large $\sigma$ on near the point (on the right) is due to the way these charges are repelled by the larger total charge residing on the left.  The mutual repulsion of the charges on the right limits how tightly they can be packed together, but the the larger repulsion from the charges on the left still manages to cram them together fairly tightly.
