Let $X$, $Y$ be differentiable manifolds, and $f : X \to Y$ a smooth surjection. Then $Y$ is said to be a quotient of $X$ if
1) $Y$ has the quotient topology 2) A function $g : Y \to \mathbb{R}$ is differentiable iff $g \circ f$ is.
One common example of this situation is a locally trivial fibration, i.e. there is a cover of $Y$ by open sets $U$ so that the restriction of f to $f^{-1}(U)$ is diffeomorphic to the projection $U \times Z \to U$, for some smooth manifold $Z$.
Lie group quotients and homogeneous spaces are examples of quotients which are locally trivial fibrations.
I don't know of an example of a quotient which is not a locally trivial fibration. Do you? It would just be a nice thing to have in mind.
(I am not insisting that the fibers be all the same, then it is trivial to construct an example just by taking disjoint unions. Anyway, I think that in a locally trivial fibration, the diffeomorphism class of the fiber is locally constant.)
Let $X$ denote $\mathbb{R}^2$ with the origin removed and let $Y$ be $\mathbb{R}$ with $f:X\rightarrow Y$ given by $f(x,y) = x$.
First, $f$ is not a locally trivial fibration. If it were, the fibers would all have the same homotopy type, but the fiber over $0$ is disconnected while the fiber over every other point is connected.
Now we show the other properties.
$f$ is a quotient map. Indeed, the projection map $\mathbb{R}^2\rightarrow \mathbb{R}$ is open and $f$ is the restrction of the projection to the open set $X\subseteq \mathbb{R}^2$.
Now, let $g:Y\rightarrow \mathbb{R}$ be any function.
Assuming $g$ is smooth, then $g\circ f$ is smooth since $f$ is smooth.
Conversely, let $i:\mathbb{R}\times \{1\} \rightarrow X$ denote the inclusion, which is obviously smooth. Then $((g\circ f)\circ i) = g(f(i(x)) = g(f(x,1))= g(x)$, so $g$ is a composition of smooth functions, so is smooth.