Now, I am aware that the divergence of a vector field, $\vec{F}$, can be defined in two ways. What I don't understand is why do these equal each other formally?
Definition 1: $$\text{div}\vec{F} = \lim_{V \rightarrow 0} \frac{1}{V} \oint{\vec{F} \cdot d\vec{A}}$$
Where $V$ is the volume of some arbitrary region which shrinks around a point $P = (x, y, z)$ and $d\vec{A}$ is the normal area element vector on a closed surface enclosing region $R$ which has volume $V$.
Definition 2: $$ \text{div}\vec{F} = \nabla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$$
Any illumination would be appreciated.
Consider a small box located at $\textbf{r} = x\hat{\textbf{i}} + y\hat{\textbf{j}} + z\hat{\textbf{k}}$. The volume of the box is $\delta V = \delta x\delta y\delta z.$ Also consider a vector field of $\textbf{F} =F_x\hat{\textbf{i}} + F_y\hat{\textbf{j}} + F_z\hat{\textbf{k}}$.
The two faces that lie in the y-z plane have unit normals $\hat{\textbf{n}} = \pm \hat{\textbf{i}}$, so that the contributions to the flux out the box from these faces is
$$(\hat{\textbf{F}}\cdot\delta \hat{\textbf{S}})_{+x} = F_x(x+\delta x, y, z)\delta y\delta z$$ $$(\hat{\textbf{F}}\cdot\delta \hat{\textbf{S}})_{-x} =-F_x(x, y, z)\delta y\delta z$$
So that the net flux in this direction is given as
$$(\hat{\textbf{F}}\cdot\delta \hat{\textbf{S}})_x = \frac{F_x(x+\delta x, y, z) - F_x(x, y, z)}{\delta x}\delta x\delta y\delta z$$
$$(\hat{\textbf{F}}\cdot\delta \hat{\textbf{S}})_x = \frac{\partial F_x}{\partial x}\delta V$$
Doing the same for the $y$ and $z$ components and summing over them to get the total flux
$$\sum_{\mathrm{All\ Faces}}\hat{\textbf{F}}\cdot\delta \hat{\textbf{S}} = \mathbf{\nabla}\cdot\hat{\textbf{F}}\delta V$$
Now this can be generalized to an arbitrary volume by breaking that volume up into small boxes of volume $\delta V$. Now the fluxes from the adjacent faces cancel, so that only the fluxes out the external faces contribute. Therefore
$$\sum_{\mathrm{ext.\ faces}}\hat{\textbf{F}}\cdot\delta \hat{\textbf{S}} = \sum_{\mathrm{all\ cubes}}\mathbf{\nabla}\cdot V\delta V$$
$$\frac{1}{\delta V}\sum_{\mathrm{ext.\ faces}}\hat{\textbf{F}}\cdot\delta \hat{\textbf{S}} = \sum_{\mathrm{all\ cubes}}\mathbf{\nabla}\cdot V$$
In the limit as $\delta V \to 0$, this gives
$$\text{div}\textbf{F} = \lim_{\delta V \to 0} \frac{1}{\delta V} \oint{\textbf{F} \cdot d\textbf{S}}$$
Hopefully this is insightful as to why both are equivalent.