How to find electric field at a point on an arbitrary surface charge $(\sigma)$ due to it? Electric field at any point due to any volume charge $(\rho)$
Electric field at a point outside a volume charge is given by Coulomb's law:
$$\vec{E}=k \iiint_V \dfrac{\rho}{r^2} (\hat{r})dV \tag1$$
To find the electric field inside a volume charge, we have to deal with the singularity. So let's use spherical coordinate system where $dV=r^2 \sin\theta\ d\theta\ d\phi\ dr$. Then equation $(1)$ becomes:
$$\vec{E}=k \iiint_V (\rho) (\hat{r}) \sin\theta\ d\theta\ d\phi\ dr$$
Here we can see that the electric field doesn't blow up even at a point inside the volume charge distribution and the singularity is removable. Therefore equation $(1)$ is applicable for finding the field even at a point inside the volume charge.
Electric field due to an arbitrary surface charge $(\sigma)$ at a point not on the arbitrary surface charge $(\sigma)$
$$\vec{E}=k \iint_A \dfrac{\sigma}{r^2}(\hat{r})dA \tag2$$
Electric field due to an arbitrary surface charge $(\sigma)$ at a point on the arbitrary surface charge $(\sigma)$
I have no clue how to deal with the singularity in this case. Any help will be appreciated.
 A: Next to the surface (i.e. a point not on the surface), there isn't a problem: your integral
$$\vec{E}=k \iint_A \dfrac{\sigma}{r^2}(\hat{r})dA \tag2$$
might have a large integrand, but $r$ is bounded away from zero, and you don't have any problems with the singularity, as the integrand (however large) is finite, continuous, and bounded.
On the integral itself, on the other hand, the electric field itself isn't well-defined, and its component along the surface normal will have a discontinuity, with a discrete jump by $4\pi k \: \sigma(\vec r)$ (modulo constants). 
This means, therefore, that if you want to assign a value of the electric field to the surface itself then you need to be very careful about how you're defining that value and what you want to use it for. The normal methods, however, leave that value unassigned.
A: 
I have no clue how to deal with the singularity in this case. Any help will be appreciated.

How to deal with a singularity in a formula depends on what we are trying to achieve. Sometimes, non-existence of value is the correct answer (e.g. electric field of a point particle at its position). Other times, one has to ditch the formula as inadequate and find the actual result in a different way.
If the task is to get value of electric field at a point on a charged surface, usually that has a definite solution, a finite value.
It is true that electric field is discontinuous in such case, but physically there is a very good reason to assign definite electric field even at the point of discontinuity.
This can be found in different ways, but I like the way which goes back to physics and the very meaning of electric field.
Let us recall how electric field of a set of charged bodies at any point of space is defined in physics:
force acting on a test particle of small charge $q$ (so it does not upset distribution of charge on the bodies) placed at the point, divided by charge $q$.
We can find electric field at any point using this definition as long as the test particle, when put there, experiences definite force.
In case of charged surface, cut out a small disk of the charged surface of finite charge $\Delta Q$ but keep it where it was by applying net force, denoted $\Delta \mathbf F$. The system is thus divided into two parts: the disk and the remaining body (perforated surface).
Now, use the disk as test particle to find electric field of the remaining body, using the definition:
$$
\mathbf E = \frac{-\Delta \mathbf F}{\Delta Q}.
$$
This would be the experimental procedure and for common cases such as charged metallic body without sharp edges we know it leads to definite value of electric field (this manifests as finite tension acting on the surface of the charged body).
If it works experimentally, it must work also in the proper calculation of electric field, one just has to exclude suitable small piece of charged matter around the point of interest. That piece of charge, due to electric forces obeying principle of action and reaction, does not contribute to the relevant force in the above formula.
For uniformly charged sphere of radius $R$, this is easy to calculate. Consider electric field at a point just above the charged surface, separated by small distance $h$, which we know is equal to $\sigma/\epsilon_0$. This field can be calculated as sum of two contributions:
1) field due to small disk of radius chosen in such a way that $r\ll R$ but even $h\ll r$;
2) field of the remaining perforated sphere.
It is known that field of a disk just above its surface is equal to $\frac{\sigma}{2\epsilon_0}$. Thus to get net field $\sigma/\epsilon_0$ at the height $h$ the field 2) has to be also equal to $\frac{\sigma}{2\epsilon_0}$.
But the value 2) is, due to continuity across the empty hole, also value of the field acting on the very disk itself, so it is the magnitude of electric field at the surface.
A: The general idea is to apply the integrals on the surface, skipping small areas around the point at which you are trying to find the field, then taking the limit as the size of the skipped areas goes to zero. 
This is referred to as finding the Cauchy principal values of improper integrals, as explained in the following reference:  https://en.wikipedia.org/wiki/Cauchy_principal_value
In formulating numerical methods called "panel methods" in potential-flow aerodynamics, surfaces are approximated by a large number of small panels, and simple functions are used to describe both the shape and the charge distributions over each panel. This allows the integrals to be found analytically over each panel, so that the singular behavior can be dealt with exactly. 
There are many successful examples of this type of numerical procedure. I believe the 1st 3D application was done by Hess & Smith. Mathematical details are given in the following report:
https://apps.dtic.mil/dtic/tr/fulltext/u2/755480.pdf
