This is not really about the connection with graph theory, a topic I am rather ignorant of, but rather the connection to continuum notions, all of which I learned from this paper.

Consider a simplicial complex in 3 dimensions for simplicity of visualization. The 0-simplexes are vertices $(i)$, the 1-simplexes are bonds $(ij)$, 2-simplexes are triangles $(ijk)$, 3-simplexes are tetrahedra $(ijkl)$. Each simplex has an orientation and under permutation of vertices acquires a sign change of +1 or -1 if the permutation is even or odd respectively.

Now we can define functions ($p$-chains) on our simplicial complex,
$$\phi = \sum_i \phi_i (i)$$
$$\alpha = \sum_{[ij]} \alpha_{ij} (ij)$$
$$\beta = \sum_{[ijk]} \beta_{ijk} (ijk)$$
$$\gamma = \sum_{[ijkl]} \gamma_{ijkl} (ijkl)$$
where the $\alpha_{ij}$ etc. are fully anti-symmetric and the sum is over equivalence classes of simplexes (i.e. we pick one representative for each simplex from its possible permutations).

Now we define a boundary operator $\partial_p$ on $p$-simplexes. On a 0-simplex, we have $\partial_0(i) = 0$. For a 1-simplex we have
$$\partial_1(ij) = (j) - (i)$$
and we generalize this,
$$\partial_p(i_0 \cdots i_{p-1}) = \sum_n (-1)^n (i_0 \cdots \hat{i}_n \cdots i_{p-1})$$
where the hat means that vertex is removed. This is equivalent to saying that the boundary of a $p$-simplex is the sum of the $p-1$-simplices which bound it, each oriented such that their "edges" are oppositely oriented. Thus for a triangle we find
$$\partial_2(ijk) = (jk) + (ki) + (ij)$$
while for a tetrahedron we have
$$\partial_3(ijkl) = (jkl) + (kli) + (lij) + (ijk)$$
This construction automatically satisfies $\partial_{p-1} \partial_{p} = 0$ due to the "oppositely oriented edges" condition above.

Next, define the coboundary operator $\partial_p^\dagger$ which takes $p$-chains to $p+1$-chains. The definition says
$$\partial_p^\dagger (i_1 \cdots i_{p}) = \sum_{i_0@[i_1 \cdots i_{p}]}
(i_0 \cdots i_{p})$$
where $@$ means "adjacent to". Thus for a 0-simplex,
$$\partial_0^\dagger (j) = \sum_{i@j} (ij)$$
Note that the sum is over oriented 1-simplices which "point towards $(j)$". For a 1-simplex $(ij)$, $\partial_1^\dagger(ij)$ is the sum is over all triangles $(i_0 i_1 i_2)$ such that $\partial_2(i_0 i_1 i_2)$ contains $+(ij)$, and so on. This operator also satisfies $ \partial_{p+1}^\dagger \partial_p^\dagger = 0$ by construction.

The boundary and co-boundary operators act on $p$-chains linearly. We can draw an analogy with differential geometry --- in particular, the co-boundary operator is analogous to the exterior derivative, and $p$-chains are akin to exterior $p$-forms. As shown in the above-linked paper, we can think of $0$-chains as scalar fields, $1$-chains as vector fields, $2$-chains as pseudo-vector fields, and $3$-chains as pseudo-scalar fields. The properties of the boundary operators are then summed up in this figure (their $d$ is my $\partial^\dagger$):

Note that the correspondence is not an approximation (see the text for details), although one can make a connection with the continuum differential operators via a Taylor-expansion approximation in the continuum limit as the lattice spacing goes to zero.

One can now define certain vector-product operations, demonstrate Stoke's theorem, etc. utilizing this construction. In particular, we can define the Laplacian for $p$-chains as
$$\Delta_p = - (\partial_{p+1}\partial_{p}^\dagger + \partial_{p-1}^\dagger \partial_p)$$
then from the figure we find the correspondence
$$\Delta_0 \sim \mathrm{div}\,\mathrm{grad} $$
$$\Delta_1 \sim \mathrm{grad}\,\mathrm{div} - \mathrm{curl}\,\mathrm{curl}$$
$$\Delta_2 \sim \mathrm{grad}\,\mathrm{div} - \mathrm{curl}\,\mathrm{curl}$$
$$\Delta_3 \sim \mathrm{div}\, \mathrm{grad}$$

In particular, $\Delta_0 = -\partial_1 \partial_0^\dagger$ is the usual graph Laplacian, and one can show (with appropriate choice of representatives in the summations above), that
$$\Delta_0 = A - D$$
where $A$ is the adjacency matrix and $D$ is the incidence matrix of the graph (see here). In coordinate notation, it looks like
$$\Delta_0 \phi = - \partial_1 \partial_0^\dagger \sum_i \phi_i (i)$$
$$ = - \partial_1\sum_{i} \phi_i \sum_{j@i} (ji)$$
$$ = - \sum_{i} \phi_i \sum_{j@i} [(i) - (j)]$$
$$ = - \sum_{i} (i) \sum_{j@i} (\phi_i - \phi_j)$$
from which it is easy to see that the above expression is correct:
$$
\Delta_0 \phi = \sum_{i} (i) \sum_{j@i} \phi_j - \sum_{i} (i) \sum_{j@i} \phi_i \\
= \sum_i (i) \sum_j (A_{ij} - D_{ij}) \phi_j
$$
where $D_{ij} = \delta_{ij} z_i$ with $z_i$ being the coordination number of vertex $i$ and $A_{ij} = \delta_{i@j}$. The higher-order Laplacian operators are then related to the graph structure of certain bond/face/body-duals of the original graph.

There is a further connection to various topics such as de Rham cohomology, the Hodge decomposition and harmonic forms. In particular, we can decompose any $p$-chain into
$$\sigma^p = \partial_{p-1}^\dagger \alpha^{p-1} + \partial_{p+1} \beta^{p+1} + \gamma^{p}$$
where $\gamma^{p}$ is a "harmonic chain" and satisfies $\Delta_p \gamma^{p} = 0$, and corresponds to a contribution which "winds around" the lattice topologically, i.e. $\gamma^{p} \in H_p$, the $p$'th homology group of the complex. I have not seen that made more explicit anywhere yet and don't know enough about the topics myself to really comment further.