Why is mean curvature extrinsic? I believe that an intrinsic property is dependent on a surface itself (not on how it is parameterized), and an extrinsic property may vary depending on the parameterization.

Why is mean curvature extrinsic then? It's a measure of the curvature of a surface, and curvature is intrinsic, right? Perhaps I'm confusing mean curvature for Gaussian curvature actually. I don't have a good intuitive grasp of what mean curvature is.
At a point on a surface, mean curvature is the mean of the two principal curvatures. The intrinsic curvature that you refer to is the Gaussian curvature, which is the product of the two principal curvatures, and can be shown to be invariant under isometry. To get a feeling for this, consider a flat piece of paper. Intuitively, it has zero mean as well as Gaussian curvature. But if you roll the paper into a cylindrical shape (let's say of unit radius), the principal curvatures are along the axis and the circular cross sections, and are $0$ and $1$ respectively, so the surface now has mean curvature $1/2$, while the Gaussian curvature is still $0$. In general, mean curvature is a property of an embedding, and depends on how the surface (or in general, manifold) is embedded into the ambient Euclidean space (or manifold). However, the Gaussian curvature depends on the surface itself (which is to say, depends solely on the metric), and does not depend on the embedding. To get more insight, you can look at the books of do Carmo or O'Neill. You will find that both the curvatures are defined in terms of the shape operator, which apriori depends on the embedding. If defined this way, neither appears to be intrinsic, but then one proves the intrinsic nature of the Gaussian curvature (this is the Theorema Egregium of Gauss).
the local values of curvature are necessarily dependent on the parametrization; what is invariant is the global property of curvature (e.g. when you integrate over the whole manifold etc.)