Orbifold of the three-sphere (and lens spaces) Think of the three-sphere as given by $\lbrace|z|^2+|w|^2=1, \;z,w\in \mathbb{C}^2\rbrace$. We can regard it in terms of Hopf coordinates
\begin{align*}
z&= \cos(\theta/2)e^{i(\phi+\psi)}\\
w&= \sin(\theta/2)e^{i\psi}
\end{align*}
where $0\leq\theta\leq \pi$ and $0\leq \phi,\psi<2\pi$. Now I want to consider the orbifold obtained when we divide $S^3$ by the following action:
$$(z,w)\mapsto (z,e^{2\pi i/k}w)$$
for $k$ some integer. In terms of the Hopf coordinates, this identifies
$$(\phi,\psi)\sim (\phi-2\pi /k,\psi+2\pi /k)$$
Is the resulting space a Seifert fibration with 
$$(b,g;(p_1,q_1),(p_2,q_2))=(1,0;(k,-1),(-k,1))$$
where $b$ is the Euler number of the circle fibration, $g$ the genus of the base and $(p_i,q_i)$ label the fixed points of the action described above?
Also, applying the same orbifolding to the Lens spaces $L(n,1)$, is it correct that the resulting orbifolds are the Seifert fibrations
$$(b,g;(p_1,q_1),(p_2,q_2))=(n,0;(k,-n),(-k,n))$$
And if this is true, are these orbifolds also known in another way than just by their Seifert "invariants"?
 A: The quotient of $S^3$ by your action is again $S^3$. You can see that by dividing the sphere into regions where (say) $|w| \leq 1/2$ and $|w| \geq 1/2$. The quotient of each of these is a solid torus, and you can check that they are glued together to give a 3-sphere. A sanity check is that the fixed point set is the circle $w=0$, and that the projection gives the standard branched cover of $S^3$ over itself. The quotient orbifold is $3$-sphere with an unknot branch set of order $k$.
I'm not sure what you mean by `applying the same orbifolding to the Lens spaces...'.  
A: Orbifolds which are the cyclic quotients of $S^3$ are discussed quite elegantly in Section 3 of (apologies the citation engine was misbehaving on me):
Boileau, M., Boyer, S., Cebanu, R. and Walsh, G.S., 2012. Knot commensurability and the Berge conjecture. Geometry & Topology, 16(2), pp.625-664.
Their approach is to discuss these orbifolds in two ways: 1) as quotients of $S^3$ and 2) as two orbifold quotients of tori glued together along their boundaries (which provides the orbifold equivalent of a Seifert fibration). The base orbifold of these fibrations is in general a bad orbifold. 
Although not relevant to the arguments of the above cited paper, but relevant to a treatment of these kinds of objects as Seifert fiber orbifolds, there are orbifold which decompose into two cyclic quotients of solid tori mentioned which are not cyclic quotient of $S^3$. These arise in two ways.
First, we can construct such an orbifold by taking two tori and quotienting out by the same cyclic  group of rotations about their cores $\mathbb{Z}/n\mathbb{Z}$. Then attaching these two "orbi-tori" via a homeomorphism $h$ of their boundaries. In this case, the corresponding orbifold has an abelian, but non-cyclic group. So long as $h$ does not identify the boundary of the disk (in the underlying space of) one orbi-torus with the other boundary of a disk, then the group will be finite abelian. In any case, the Sefiert fibration will be over a base orbifold $S^2(p,q)$ where $p,q$ are not necessarily relatively prime. 
If however, we allow the homeomorphism $h$ to identify to two boundaries of disks while also building the orbifold from two orbi-tori: $S^1\times B^2/\mathbb{Z}/m\mathbb{Z}$ and $S^1\times B^2/\mathbb{Z}/n\mathbb{Z}$, then resulting orbifold will be covered by $S^1xS^2$. Consequently, it has infinite $\pi_1$. 
