Why do charges accumulate at the surface and at the the tip of the needle?

The charges distribute on the surface because this is the only place where a stable equilibrium is possible. If you had a charge inside the conductor, it would get pushed by the other charges until it cannot be pushed anymore, which happens when it meets the surface of the material.

The reason why the charge density is largest at the tip can be seen from the following example:

Consider two spheres, a large one with radius $R$ and a small one with radius $r$. They are connected with a thin wire and contain a negative excess charge ($Q$ and $q$ on the big and small spheres, respectively). Because they are connected by the wire, the electrostatic potential of the spheres must be equal in equilibrium. Otherwise charges would just move to the sphere with the lower potential. Thus we can equate the two potentials:

$$ frac{q}{r} = frac{Q}{R} $$

From this we see that the ratio of the charges equals the inverse ratio of the radii.

$$ frac{q}{Q} = frac{r}{R} $$

Hence the electric field (and equivalently, the charge density) at the surface of the smaller sphere is larger then the field at the surface pf the larger one:

$$ E_{small}=frac{q}{r^2}=E_{big} frac{R}{r}$$

You can think about the tip of the needle as the small sphere in our example. This shows that the charge density will be largest at the pointy places of a conductor.