Given a topological space $M$ with a locally free $S^1$ action on it, assume the slice representation holds,(this is often the case, e.g. M is a smooth manifold) then this will make $M$ a principal $S^1$-orbibundle over $M/S^1$. What's the proper way to define the Euler class of this $S^1$-orbibundle? How should the naturality of Euler class be stated in the category of $S^1$-orbibundles?

Here is a topological construction of such a class, in singular cohomology with rational coefficients.

Let $M$ be an $S^1$-space. Then there is the Borel construction $M // S^1 := ES^1 \times_{S^1} M$. It comes with a map $f$ to $BS^1= CP^{\infty}$ which is a fibre bundle with fibre $M$. Moreover, there is a map $q: M//S^1 \to M/S^1=Q$.

As a warm-up, let me give an alternative construction of the Euler class in the free case, i.e. for principal bundles. In that case, $q$ is a homotopy equivalence. Now put $e:= (q^{\ast})^{-1} f^{\ast} (z) \in H^2 (Q)$, where $z \in H^2 (BS^1)$ is the Euler class of the universal bundle. Then $e$ is the Euler class of the bundle $M \to Q$.

To prove this, use universality to reduce to the case of $S^1$ acting on $ES^1$. In that case, both $q$ and $f$ are homotopy equivalences. There is a sign issue here, but allow me ignore it.

Your question was of course about nonfree actions. The step that fails is that $q$ is an isomorphism in integral cohomology, because it is a homotopy equivalence. However, if the action is locally free (and everything else is nice), then $q$ is still a rational cohomology isomorphism.

This can be seen by several methods; one method is similar to the argument I sketched in my answer to this question Euler characteristic of orbifolds. The idea is that the fibre of $q$ over a point of $M/S^1$ with stabilizer group $G$ is $BG$, which has the rational cohomology of a point. The niceness of the action is needed to promote this observation to small open sets, i.e. to find a cover of $Q$ by open sets $U$ such that $q^{-1}(U) \to U$ is a rational isomorphism. Then apply Mayer-Vietoris to globalize.

Once $q^{\ast}$ is shown to be an isomorphism with rational coefficients, it can be inverted and you can apply the construction from the first part of the answer. Naturality is clear.

Since the base space is no longer a manifold but a stratified pseudomanifold, the correct way to define the Euler class may be as a two dimensional class in the intersection cohomology, as in this paper.

On the other hand there is a definition of characteristic classes of orbifold vector bundles in orbifold (Chen-Ruan) cohomology, which can be found here.