Perhaps a more geometric way of viewing mean curvature (at least for a hypersurface) is as follows:

If $\varphi: \Sigma^n\hookrightarrow (M^{n+1},g)$ is an embedded (oriented) hypersurface, then we can naturally extend a family of embeddings by choosing a unit normal $\nu$ and flowing by unit speed, i.e.

$$\varphi_t(x) := \exp_x(t \nu(x))$$

For small enough $t \in (-\epsilon,\epsilon)$, this still gives an embedding $\varphi_t: \Sigma\to M$. Now, one could pull back the metric $g$ by $\varphi$ to obtain a metric on $\Sigma$. This metric gives a (Riemannian) volume form on $\Sigma$, $d\mu_t$. Then,

$$ \frac{\partial}{\partial t} d\mu_t = H d\mu_t $$

(depending on the normalizations--I think this is not quite the same as in the other answers, but you're asking for intuition..)

This follows from the first variation formula.

In other words, the mean curvature measures how the volume of the submanifold (locally) changes under flowing the surface in the direction of the unit normal.

Perhaps a more geometric way of viewing mean curvature (at least for a hypersurface) is as follows:

If $\varphi: \Sigma^n\hookrightarrow (M^{n+1},g)$ is an embedded (oriented) hypersurface, then we can naturally extend a family of embeddings by choosing a unit normal $\nu$ and flowing by unit speed, i.e.

$$\varphi_t(x) := \exp_x(t \nu(x))$$

For small enough $t \in (-\epsilon,\epsilon)$, this still gives an embedding $\varphi_t: \Sigma\to M$. Now, one could pull back the metric $g$ by $\varphi$ to obtain a metric on $\Sigma$. This metric gives a (Riemannian) volume form on $\Sigma$, $d\mu_t$. Then,

$$ \frac{\partial}{\partial t} d\mu_t = nH d\mu_t $$

(I think this matches the normalization in @Joseph O'Rourke's answer, but of course the sign changes with the choice direction of the normal)

This follows from the first variation formula.

In other words, the mean curvature measures how the volume of the submanifold (locally) changes under flowing the surface in the direction of the unit normal.