On the electric field created by a conductor The electric field created by a conductor at a point $M$ extremely close to it is $\vec{E}=\vec{E_1}+\vec{E_2}$ where $\vec{E_1}$ is the electric field created by such a tiny bit of the conductor that we can suppose it to be a plane, and since $M$ is extremely close to the conductor such that the distance is really small compared to the size of the plane we further ahead assimilate it to an infinite plane and hence $\vec{E_1}=\frac{\sigma}{2\epsilon_0}$ and this is where I block, when we use Gauss' law on an infinite plane we also account for the electric fields on the other side of the cylinder (here our gaussian surface), but in the case of the conductor the electric field inside of it would be $\vec{0}$ and so $\vec{E_1}$ should be $\frac{\sigma}{\epsilon_0}$.
I cannot see where I've gone wrong.
 A: You are not alone about being confused about this topic and in part is is because of the use the same symbol $\sigma$ being used to mean two different things; sheet charged density and surface charge density.  
On the HyperPhysics website there is a derivation Electric Field: Sheet of Charge as shown below  
 
The sheet charge density $\sigma$ is related to the total charge residing on both surfaces of a piece of conducting sheet not the charge residing on one surface of a piece of conducting sheet.  
Note that $\sigma$ has not been called the surface charge density in the HyperPhyics derivation.

Let me change the definition of a symbol.
In the diagram below the sheet charge density is $\Sigma$ per unit area.

So the total charge on the sheet (with charges residing above and below the sheet) is $\Sigma A$.  
In this case the surface charge density is $\sigma = \dfrac{\Sigma A}{2A} = \dfrac {\Sigma}{2}$

In your example you are dealing with only one surface which has a surface charge density which is double the surface charge density that was used in the HyperPhysics derivation and so you should expect the electric field to be twice as large.
