The $\frac12$ is just a multiple to make things seem nicer, as it cancels out the square power if you differentiate the cost function $\|X-M\|^2$ with respective to the entries of $X$. Since $\|X-M\|^2$ and $\frac12\|X-M\|^2$ share the same minimizers, it's just a matter of taste to multiply by one-half.
By the way, I would call $\min_{\textrm{rank}(X)\le r}\|X-M\|^2$ a low-rank approximation problem rather than a matrix completion problem. To my understanding, a matrix completion problem is one that you need to fill in some missing entries of $M$, or modify some zero entries of $M$, so that the resulting matrix satisfies some requirement. In $\min_{\textrm{rank}(X)\le r}\|X-M\|^2$, the requirements are placed on $X$ rather than $M$, and existing nonzero entries of $M$ may change. So I wouldn't call it a "matrix completion problem". Yet different research communities have different usages of the term, so this is just my own opinion.