If $f$ is an immersion and $g$ is a submersion, then is $g \circ f $ a local diffeomorphism? I don't think so; the counter example I had in mind was $f : \mathbb{R}^2 \to \mathbb{R}^3 , f(x,y) = (x,y,x)$ and $g:\mathbb{R}^3 \to \mathbb{R}^2, g(x,y,z) =(x-z, y-z)$. Is my example right?
 A: The differential of $f$ (at any point of $\mathbb{R}^2$) has standard matrix 
$$\left[\begin{array}{cc} 1 & 0\\ 0 & 1\\ 1 & 0\end{array}\right].$$ 
As this matrix has trivial null space, the differential is injective at every point of $\mathbb{R}^2$ so $f$ is an immersion.
The differential of $g$ (at any point of $\mathbb{R}^3$) has standard matrix 
$$\left[\begin{array}{ccc} 1 & 0 & -1\\ 0 & 1 & -1\end{array}\right].$$ 
As this matrix has rank two, the differential is surjective at every point of $\mathbb{R}^3$ so $g$ is an submersion.
However, the differential of $g\circ f$ (at any point of $\mathbb{R}^2$) has standard matrix
$$\left[\begin{array}{ccc} 1 & 0 & -1\\ 0 & 1 & -1\end{array}\right]\left[\begin{array}{cc} 1 & 0\\ 0 & 1\\ 1 & 0\end{array}\right] = \left[\begin{array}{cc} 0 & 0\\ -1 & 1\end{array}\right].$$
As this matrix is singular, $g\circ f$ is not a local diffeomorphism. So your example does work.

Note, if $f : L \to M$ is an immersion and $g : M \to N$ is a submersion, then as you've demonstrated, $g\circ f$ may not be a local diffeomorphism. However, if one adds the condition that $L$, $M$, and $N$ have the same dimension, then $g\circ f$ is a local diffeomorphism (because $f$ and $g$ are local diffeomorphisms).
