Let G be a Lie group acting on a manifold M And say that there exists a point $m \in M$ such that the stabilizer of the point is non trivial. I am unable to understand how the quotient M/G fails to be a manifold. More precisely, how is a singularity created at the point $m \in M$ while constructing the quotient. Can anybody help me understand this?



Examples of Lie group actions on manifolds with singular quotients
A general nonsense answer to your question would be the orbit decomposition theorem. Given a compact lie group $G$ acting on a manifold $M$, it describes a stratification of $M$ by orbit types. 


The main theorem is that if $p \in M$, and $G\_p$ is the stabilizer of $p$, then there is a $G$-equivariant diffeomorphism between a neighbourhood of $G.p$ (the orbit of $p$) in $M$ and 


$$V \times\_{G\_p} G$$


where $V$ is a vector space and $G\_p$ acts linearly on $V$, i.e. there is a representation $G \to Aut(V)$. So $V \times\_{G\_p} G$ denotes the quotient of $V \times G$ by the diagonal action of $G\_p$. $V \times\_{G\_p} G$ is an equivariant tubular neighbourhood of $G.p$, and the projection map $V \times\_{G\_p} G \to G.p$ is simply projection onto the 2nd factor $V \times\_{G\_p} G \to G/G\_p$ followed by the identification $G/G\_p \equiv G.p$ given by the action. 


So the underlying reason for $M/G$ to have singularities is that $(V\times\_{G\_p} G)/G \simeq V/G\_p$ i.e. the action of $G\_p$ on $V$ tends to have a singular quotient. Igor gave an example of this in his comment. 


