= Solution
An <abstract smooth surface> is a Hausdorff second-countable topological space with an atlas of <manifold chart>[charts] to open subsets of $\mathbb R^2$ whose transition maps are smooth. It is <orientable smooth manifold>[orientable] when it has such an atlas for which every transition map has positive <Jacobian determinant>. A map $f:S_1\to S_2$ is a <smooth map between manifolds>[smooth map] when every coordinate representation
$$
\psi\circ f\circ\phi^{-1}
$$
is smooth wherever it is defined.
The map $a:C\to C$ is a smooth involution with no fixed point: the equation
$$
(x,y,z)=(-x,-y,-z)
$$
would force $x=y=0$, which is impossible on $x^2+y^2=1$. For each $p\in C$, choose a sufficiently small coordinate neighbourhood $U_p$ such that
$$
U_p\cap a(U_p)=\varnothing.
$$
Then the quotient projection restricts to a homeomorphism
$$
\pi|_{U_p}:U_p\longrightarrow \pi(U_p).
$$
Transporting a smooth chart $\phi_p:U_p\to\mathbb R^2$ across this homeomorphism gives the quotient chart
$$
\widetilde\phi_p
=\phi_p\circ(\pi|_{U_p})^{-1}.
$$
On an overlap, a lift lies either in another chosen neighbourhood or in its image under $a$. The corresponding transition map is therefore a transition map on $C$, possibly composed with the diffeomorphism $a$, and is smooth. Since this is a free action of the finite group $\{1,a\}$, the quotient is Hausdorff and second countable. This constructs the <smooth quotient by a free finite group action>, and in these charts $\pi$ is locally the identity. Hence $\pi$ is a <local diffeomorphism>, in particular smooth.
To test orientability, parametrize the cylinder by
$$
F(\theta,z)=(\cos\theta,\sin\theta,z).
$$
The involution acts in these coordinates as
$$
(\theta,z)\longmapsto(\theta+\pi,-z),
$$
whose derivative has determinant $-1$. Thus $a$ is an <orientation-reversing diffeomorphism> of the cylinder.
If the quotient surface $S$ were orientable, its orientation would pull back through the local diffeomorphism $\pi$ to an orientation of $C$. The identity $\pi\circ a=\pi$ would then force $a$ to preserve that pulled-back orientation, contradicting the negative determinant above. Therefore
$$
\boxed{S\text{ is not orientable}}.
$$
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