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.
 A: What struck me about your question is the sentence "curvature is intrinsic" which is just not true. There are different kinds of curvature, some are extrinsic, some are intrinsic. Intrinsic curvatures are all constructed from the Riemann tensor, which is intrinsic; extrinsic curvatures are constructed from the second fundamental form of the surface (which depends on the embedding).
For example, a circle in the plane is obviously curved, but you wouldn't notice when moving inside the circle. All one-dimensional manifolds have zero intrinsic curvature. 
The Gauss curvature and the Mean curvature are both constructed from the second fundamental form (if we are speaking about a Hypersurface immersed in euclidean space). The first is its determinant, the second its trace. The fun part is that the Gauss curvature turns out not to depend on the immersion and be intrinsic in fact (it can be constructed from the Riemann tensor as well).
So the question should in fact be: Why is Gaussian curvature intrinsic?
A: 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).
A: I felt I could offer a bit more rigor. What does it mean for something to be intrinsic? It simply means that the expression can be expressed entirely in terms of the 1st fundamental form $I$. If we use $\phi:U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$ as our parameterization, then at a point $p \in S$ and a tangent vector $w=a\phi_1 +b\phi_2 \in T_pS$, we can write $I_p(w)=g_{11}a^2+2g_{12}ab+g_{22}b^2$, where $g_{ij}=\phi_i \cdot \phi_j$. Note that $g_{11}=I_p(\phi_1)$, $g_{22}=I_p(\phi_2)$, and $g_{12}=\frac{I_p(\phi_1+\phi_2)}{2}$. Indeed it follows any expression written in terms of $g_{ij}$ implies the expression is also written in terms of $I$. If you look up the Brioschi formula, you can see that indeed Gaussian curvature is entirely expressible in terms of $g_{ij}$, and consequently in terms of $I$.
For mean curvature, we can't do this. No matter how clever you are, any expression that is written in terms of $g_{ij}$ will never equal mean curvature. 
Also note that being intrinsic to a surface does not imply being invariant under parameterization, despite the similar names. For instance, $g_{11}+g_{22}$ is intrinsic to the surface, but the expression is clearly not invariant under parameterization.
Why does it matter if a quantity is intrinsic to a surface? It sorta means if you were a bug on this surface small enough that the surface is perfectly flat to you, you'd still be able to measure that quantity. You can measure area, length, and angles. You can also measure Gaussian curvature if you're clever enough (there are formulae that involve measuring the area of certain geodesic circles).
Suppose I want to measure the rate of change of the Gauss map $N$ with respect to my trajectory. I'm a little bug but I know my trajectory $\alpha(t):[a,b] \rightarrow S$. At a point $t \in [a,b]$, I am going $\alpha'(t)$ meters per second. Great, I'm able to know that. Heck we do it all the time with cars. Now how do I figure out $(N \circ \alpha)'(t)=dN_{\alpha(t)}(\alpha'(t))$...err..I have to find $\lim_{h \rightarrow \infty }\frac{N(\alpha(t+h))-N(\alpha(t))}{|h|}$...but this requires measuring the rate of change of a vector ($N(\alpha(t)) \in \mathbb{R}^3$. $\mathbb{R}^3$!?!?! What the heck is $\mathbb{R}^3$?? I'm a tiny flat bug, whose whole world is simply this flat-but-not-really-flat surface that has these weird Euclidean-but-not-really-Euclidean properties. 
Ah, perhaps I am not clever enough. Perhaps a smarter bug can use clever higher-dimensional math to determine the rate of change of the normal vector. Well as it turns out, you can't. No matter how clever you are. You need $\mathbb{R}^3$ to figure out that rate of change. Mean curvature is defined in terms of the rate of change of the Gauss map.
A: 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.)
