Is the boundary $\partial S$ analogous to a derivative?

The surface area $|\partial S|$ of a (bounded, smooth) body $S$ is the derivative of the volume $|S_r|$ of the $r$-neighbourhoods $S_r$ of $S$ at $r=0$:

$$ |\partial S| = \frac{d}{dr} |S_r| |_{r=0}.$$

Thus, for instance, the boundary $\partial D_r$ of the disk $D_r$ of radius $r$ has circumference $\frac{d}{dr} (\pi r^2) = 2\pi r$.

More generally, one intuitively has the Newton quotient-like formula

$$ \partial S = \lim_{h \to 0^+} \frac{S_h \backslash S}{h};$$

the right-hand side does not really make formal sense, but certainly one can view $S_h \backslash S$ as a $[0,h]$-bundle over $\partial S$ for $h$ sufficiently small (in particular, smaller than the radius of curvature of $S$).

In a similar spirit, one informally has the "chain rule"

$$ {\mathcal L}_X S "=" (X \cdot n) \partial S $$

for the "Lie derivative" of $S$ along a vector field $X$, where $n$ is the outward normal. (There may also be a divergence term, depending on whether one is viewing $S$ as a set, a measure, or a volume form.) Again, this does not really make formal sense, although Stokes' theorem already captures most of the above intuition rigorously (and, as noted in the comments, Stokes' theorem is probably the clearest way to link boundaries and derivatives together).

EDIT: A more rigorous way to link boundaries with derivatives proceeds via the theory of distributions. The weak derivative $\nabla 1_S$ of the indicator function of a smooth body $S$ is equal to $-n d\sigma$, where $n$ is the outward normal and $d\sigma$ is the surface measure on $\partial S$. (This is really just a fancy way of restating Stokes' theorem, after one unpacks all the definitions.) This can be used, for instance, to link the Sobolev inequality with the isoperimetric inequality.

In a similar spirit, $\frac{1_{S_h} - 1_S}{h}$ converges in the sense of distributions as $h \to 0$ to surface measure $d\sigma$ on $\partial S$, thus providing a rigorous version of the intuitive difference formula given previously.

A partial answer to Q1 -- apologies if this is obvious, but I don't see it written here yet, and this is the thing that made me sit up and take notice of the fact that there's some sort of connection between the boundary operator $\partial$ and differentiation. If $X$ and $Y$ are two topological spaces and $A \subset X$, $B\subset Y$ are closed, then they satisfy a product rule of sorts: $$ \partial(A\times B) = ((\partial A)\times B) \cup (A \times (\partial B)). $$ This also works without the assumption that $A$ and $B$ are closed if you're willing to weaken the analogy a bit by replacing the right-hand side with $\partial(A\times B) = ((\partial A)\times \overline{B}) \cup (\overline{A} \times (\partial B))$.

I like Vaughn's answer, but it seems to me that another form of product rule holds. If $A$ and $B$ are reasonnable subsets of $\mathbb R^n$ (perhaps one needs only that they are the closure of their interior), then $$\partial(A\cap B)=(\partial A\cap B)\cup(A\cap\partial B).$$ Here, $\cap$ and $\cup$ are the boolean operators, which are the analogues of $\times$ and $+$.