Why  is this a local constant sheaf If a group $G$ acts on a topological space $M$, and a representation of $G$ on a vector space $V$, why $M \times_G V$ is a local constant sheaf over $M/G$?
 A: Dear HYYY, 
You should assume that the group $G$ acts freely and properly discontinuously on $M$.  Also,
you should equip $V$ with the discrete topology.  (This way you will get a local system rather than a vector bundle, which was a point of confusion in the comments above.)
If $G$ acts freely and properly discontinuously, then if $m$ is any point of $M$, it has a neighbourhood $U$ such that $Ug$ is disjoint
from $U$ for all non-trivial $g \in G$.  (Here I am writing the $G$-action on $M$ on
the right, as is implicit in the question.)
Thus $U$ maps injectively into $M/G$, and $U \times V$ maps injectively into
$M\times_G V$.  Hence $M\times_G V$ is locally constant (because, as we have just shown,
its pull-back over the open subset $U$ of $M/G$ is constant).
A: For a bit of intuition, consider $V = \mathbb{C}^2$ and $M = \mathbb{C}$ with $G = \mathbb{Z}/2$ acting diagonally as $\pm 1$ in both cases.
Then for any open set $U$ not containing the origin in the base, the sheaf is just the constant sheaf $\mathbb{C}^2$. If, however, $U$ does contain the origin, then it is the constant sheaf $\mathbb{C}^2/\pm 1$, and so this sheaf is locally constant.
In general, the fibre over a point $[x]$ in the base should be a copy of $V/stab_G(x)$. You should be able to study these to obtain your result.
