In $R^{n+1}$, suppose $M$ is smooth embedded hypersurface which is isomorphic to $S^{n}$. Is that the same when we talk about $M$ is mean convex (the normal vector, pointed inward), and $M$ is convex, ( in Huisken's sense）？

It's very different! Convex (in the Huisken sense) means that *all* the principal curvatures $\lambda_1,\ldots,\lambda_n$ are positive, i.e. $\lambda_1>0,\ldots,\lambda_n>0$. Mean convex only means that the mean curvature vector points inwards, i.e. just that the sum $\lambda_1+\ldots+\lambda_n$ is positive, which is obviously a much weaker curvature condition (e.g. $n-1$ principal curvatures could be negative as long as $\lambda_n$ is sufficiently positive).

Correspondingly the mean curvature flow behaves very differently. Huisken proved in 82 that convex hypersurfaces converge to a round point. On the other hand, in the mean convex case many interesting singularities can form. It's a deep theorem due to White that they are in fact all modeled on shrinking cylinders $S^j\times R^{n-j}$ (see also my paper with Kleiner for a more elementary proof). The neckpinch in $R^3$ described in Will Jagy's answer is the simplest example for such a singularity.

Huisken did not define the notion of mean convexity. A mean convex surface that is not convex could be, for example, the dumbbell shape made this way: draw an oval by making two parallel line segments of the same length, then gluing semicircles at both ends. Now gently bend the line segments a bit inward, so it is still a smooth continuous curve but not convex. Rotate this around an axis parallel to the original straight lines, the result is mean convex.

Huisken did compare mean curvature flow on convex surfaces and then mean convex. i am pretty sure he is the one who settled the nature of pinching off; in finite time, the dumbbell shape described pinches off and becomes two surfaces. Just before that happens, is narrow tube that jojns them getting closer to a cylinder, or does it pinch off as a double cone? I don't remember the answer, but he did it.