Just to add an (in my opinion important) piece of information. Say $F$ is a function on the vertices of a graph, so $F:V \to \mathbb{R}$. Then $\nabla F$ is a function from the edges to the vertices (here I see an edge as a pair of vertices $(x,y)$, so edges are oriented): $$\nabla F (x,y) := F(y) - F(x)$$ Now this definition is very natural in many ways. For example, you would expect that the integral of the gradient of a function along a path is just the difference of the values of the function at the end of this path. And this holds here: if $\vec{p}$ is an oriented path (say from $a$ to $b$) then $\sum_{\vec{e} \in \vec{p}} \nabla F(\vec{e}) = F(b) - F(a)$. You can add a weight to the edges, but this is (in my opinion) not the important point for intuition.
Here is the important piece of information: if your graph has bounded degree$^*$, $\nabla$ defines an operator from $\ell^2V$ to $\ell^2E$. (The pairing on $\ell^2V$ is just $\langle f \mid g \rangle_V = \sum_{v \in V} f(v)g(v)$. Same pairing on $\ell^2E$ just that the sum is over the edges) So you may ask, what is the adjoint of this operator? Well the defining property can be tested on Dirac masses (which are a nice basis of our spaces): $$ \langle \nabla^* \delta_{\vec{e}} \mid \delta_x \rangle = \langle \delta_{\vec{e}} \mid \nabla\delta_x \rangle $$ So this is $+1$ if $\vec{e}$ has $x$ as target, $-1$ if $x$ is the source and 0 otherwise. Extended by linearity this gives: (here $G(x,y)$ is a function on the edges) $$ \nabla^* G(x) = \sum_{y \sim x} G(x,y) - \sum_{y \sim x} G(y,x) $$ where $y \sim z$ means $y$ is a neighbour of $x$. (If your edges are not oriented, it is natural to consider only alternating functions on the edges, that is $G(x,y) = -G(y,x)$; the above expression simplifies then a bit)
The rest is just a computation: $$ \begin{array}{rl} \nabla^* \nabla F(x) &= \displaystyle \sum_{y \sim x} \nabla F(x,y) - \sum_{y \sim x} \nabla F(y,x) \\ &= \displaystyle \bigg( \sum_{y \sim x} [F(y) - F(x)] \bigg) - \bigg( \sum_{y \sim x} [F(x) - F(y)] \bigg) \\ &= \displaystyle 2 \bigg( \sum_{y \sim x} [F(y) - F(x)] \bigg) \\ &= \displaystyle 2 \bigg( \big[ \sum_{y \sim x} F(y) \big] - \deg(x) F(x) \bigg) \\ \end{array} $$ And that's the formula for the Laplacian (when the conductance is 1). Note that I got a difference of a factor of 2 (because my definition of divergence is a bit different). But having a divergence which is the adjoint of the gradient, is a very important point, in my opinion.
If you add a weight to the edges, the computation are slightly more complicated, but it's just [possibly painful] bookkeeping.
$^*$ if you have weighted edges you could have an infinite number of edges as long as their weight is bounded