Do this experiment: draw a curve in 2D and about some point on that curve, draw a unit normal vector field. Now convince yourself that if you push the curve out along the normal field by a distance $\epsilon$, the length changes by a factor of $1+\epsilon\kappa$ where $\kappa$ is the curvature.

Now consider a 2D surface in 3D. Choose local coordinates so that the first two basis vectors are tangent to the surface and the curvatures along the first two coordinates are the principal curvatures. The infinitesimal change in surface area is $(1+\epsilon\kappa_1)(1+\epsilon\kappa_2)$ -- actually this product is the Jacobian of the normal map, $N_\epsilon$ that pushes the surface out along the normal field a distance epsilon.

In general, for co-dimension one surfaces, the Jacobain of the normal map $N_\epsilon$ is $\Pi_{i=1}^{n-1} (1+\epsilon\kappa_i)$. We can integrate this over the original surface to get the new, n-1 volume of the pushed surface. That is:

$\mathcal{H}^{n-1}(N_\epsilon(W)) = \int_{W} \Pi_{i=1}^{n-1} (1 + \epsilon\kappa_i) d\mathcal{H}^{n-1}$

$ \hspace{1in} = \int_{W} 1 d\mathcal{H}^{n-1} + \epsilon \int_{W} \sum_i \kappa_i d\mathcal{H}^{n-1} + ... + \epsilon^k \int_{W} \sum_{s\in S(k)}\Pi_{i\in s}\kappa_i d\mathcal{H}^{n-1}$

$ \hspace{1.2in} + ... + \epsilon^{n-1} \int_{\partial W} \Pi_{i=1}^{n-1} \kappa_i d\mathcal{H}^{n-1}$

Now we note that the first order term is the integral of the mean curvature. That is, to first order, the change in surface volume is given by the mean curvature. Note that I am using the term mean curvature for what is sometimes called total mean curvature, $\sum_{i=1}^{n-1} \kappa_i$.

(This works in co-dimension $k>1$, but then, because the set of normal directions at any point is $k$-dimensional, it is a bit more involved to get a result that looks like the result here.)