Separating the potential energy of a system of particles. Assuming all forces derive form a conservative source and that all forces observe the strong form of the third law, how do we arrive at the following equation? 
\begin{equation}
V=\sum _i V_i+\frac 12 \sum _i \sum _{j,j\neq i}V_{ij}
\end{equation}
Okay so here are my thoughts. 
Firstly we divide the work up into internal and external components:
\begin{equation}
\sum _i\int ^{r_2}_{r_1}\vec{F_i}^{(e)}\cdot d\vec s _i+\sum _{i,j}\int ^{r_2}_{r_1}\vec {F_{ij}}\cdot d\vec s _i
\end{equation}
The factor of a half comes in since we are summing over both $i$ and $j$, and (I think) we can assume that $V_{ij}=V_{ji}$ (why?).
let,
\begin{equation}
\vec F_i ^{(e)}=-\nabla_iV_i(\vec r_1,...,\vec r_n)\\
\vec F_{ij}=-\nabla_{ij}V_{ij}(|\vec r_i -\vec r_j|)
\end{equation}
The $|\vec r_i -\vec r_j|$ is just the magnitude of the particles separation. 
Now the work is given by:
\begin{equation}
W=\sum _i\int ^{r_2}_{r_1}-\frac{\partial }{\partial \vec r_i}V_i\cdot d\vec r_i+\frac 12 \sum _{i,j}\int^{r_2}_{r_1}-\frac{\partial }{\partial \vec r_{ij}}V_{ij}\cdot d\vec r_{ij}
\end{equation} 
\begin{equation}
W=-\sum _i\int^{V_2}_{V_1}dV_i-\frac 12 \sum _{i,j}\int ^{V_2}_{V_1}dV_{ij}
\end{equation}
So my question is how we got to the end step? This is all out of Goldstein in the 1st chapter. Unfortunately I can't follow the derivation at all ...
 A: 
Firstly we divide the work up into internal and external components:
  \begin{equation}
\sum _i\int ^{r_2}_{r_1}\vec{F_i}^{(e)}\cdot d\vec s _i+\sum _{i,j}\int ^{r_2}_{r_1}\vec {F_{ij}}\cdot d\vec s _i
\end{equation}
  The factor of a half comes in since we are summing over both $i$ and $j$, and (I think) we can assume that $V_{ij}=V_{ji}$ (why?).

They don't have to be equal. They just have to be equal to within an arbitrary constant. Adding an arbitrary constant to a potential doesn't make a bit of difference (at least in classical mechanics), so one might as well call that constant zero.
To see why they have to be equal (to within a constant), assume $V_{ij} = V_{ji} + \Delta V_{ij}$, where $\Delta V_{ij}$ is some non-constant function. Then $\vec F_{ij} = -\vec\nabla_{ij} V_{ij}$ $= -\vec\nabla_{ij} V_{ji} - \vec\nabla_{ij} \Delta V_{ij}$. Note that $\vec\nabla_{ij} V_{ji} = -\vec\nabla_{ji}V_{ji} = \vec F_{ji}$ Thus $\vec F_{ij} = - \vec F_{ji} - \vec\nabla_{ij} \Delta V_{ij}$. The only way this can satisfy Newton's third law is if $\vec\nabla_{ij} \Delta V_{ij} = 0$. This contradicts the assumption that $\Delta V_{ij}$ is some non-constant function.

let,
  \begin{equation}
\vec F_i ^{(e)}=-\nabla_iV_i(\vec r_1,...,\vec r_n)\\
\vec F_{ij}=-\nabla_{ij}V_{ij}(|\vec r_i -\vec r_j|)
\end{equation}
The $|\vec r_i -\vec r_j|$ is just the magnitude of the particles separation.

You have a misunderstanding here with respect to the external forces. The force $\vec F_i^{(e)}$ and hence the potential $V_i$ depends only on $\vec r_i$. There is no dependence on $\vec r_j$ where $j \ne i$. That first line should be $\vec F_i^{(e)} = -\nabla_i V_i(\vec r_1)$. 

Now the work is given by:
  \begin{equation}
W=\sum _i\int ^{r_2}_{r_1}-\frac{\partial }{\partial \vec r_i}V_i\cdot d\vec r_i+\frac 12 \sum _{i,j}\int^{r_2}_{r_1}-\frac{\partial }{\partial \vec r_{ij}}V_{ij}\cdot d\vec r_{ij}
\end{equation} 
  \begin{equation}
W=-\sum _i\int^{V_2}_{V_1}dV_i-\frac 12 \sum _{i,j}\int ^{V_2}_{V_1}dV_{ij}
\end{equation}
  So my question is how we got to the end step? This is all out of Goldstein in the 1st chapter. Unfortunately I can't follow the derivation at all ...

Goldstein uses $\int_1^2$, not $\int_{r_1}^{r_2}$ or $\int_{V_1}^{V_2}$ This may be part of your problem. Your own notation may be confusing you.
The second line follows from the first as a consequence of the fact that $dV = \vec{\nabla} V \cdot d\vec r$ for some potential $V$.
