Covering transformation and orientation ( in Synge Theorem ) Case 1: $M_1$ is a compact orientable manifold.  And
$$
\pi_1:\tilde M_1\rightarrow M_1
$$
is the universal covering of $M_1$. Introduce on $\tilde M_1$ the orientation and metric such that $\pi_1$ preserves  orientation and is local isometry. Then how to show the covering transformation of $\tilde M_1$
$$
k_1: \tilde M_1 \rightarrow \tilde M_1  ~~~(\text{satisfy }  \pi_1\circ k_1 = \pi_1 )
$$
preserves the orientation ?
Case 2: $M_2$ is compact and not orientable. $\pi_2:\overline M_2 \rightarrow M_2$ is the orientable double cover of $M_2$. Introduce on $\overline M_2$ the covering metric. Then, how to show the covering transformation of $\overline M_2$
$$
k_2 :\overline M_2\rightarrow \overline M_2  ~~~(\text{satisfy }  \pi_2\circ k_2 = \pi_2 )
$$
reverses the orientation of $\overline M_2$ ?
PS: This problem is from the proof of Synge theorem of do Carmo's Riemannian Geometry. I try to write it in my words. But I afraid of losting anything. Therefore, I add the proof in below as picture. And the two red lines correspond the two cases.



 A: Case $1$: As $\pi_1\circ k_1 = \pi_1$, for any $m \in \tilde{M}_1$, we have
$$(d\pi_1)_m = (d(\pi_1\circ k_1))_m = (d\pi_1)_{k_1(m)}\circ(dk_1)_m.$$
As $\pi_1$ is orientation-preserving by construction, the linear maps $(d\pi_1)_m : T_m\tilde{M}_1 \to T_{\pi_1(m)}M_1$ and $(d\pi_1)_{(k_1(m))} : T_{k_1(m)}\tilde{M}_1 \to T_{\pi_1(k_1(m))}M = T_{\pi_1(m)}M$ are orientation-preserving, and hence so is $(dk_1)_m : T_m\tilde{M} \to T_{\pi_1(m)}M$. Therefore $k_1$ is an orientation-preserving diffeomorphism.
Note that we didn't need compactness here, or that the covering is the universal covering.
Case $2$: If $k_2$ is orientable, then one can define an orientation on $M_2$ as follows.
Let $U$ be an evenly covered open set, then $\pi_2^{-1}(U) = U_1\sqcup U_2$ where $\pi_2 : U_i \to U$ are diffeomorphisms; let $s_i^U : U \to U_i$ be the inverse diffeomorphisms. Note that $k_2$ restricts to a self-diffeomorphism of $U_1\sqcup U_2$ which interchanges $U_1$ and $U_2$, and since $\operatorname{id}_U = \pi_2\circ s_1^U = \pi_2\circ k_2\circ s_1^U$, we have $s_2^U = k_2\circ s_1^U$ and likewise $s_1^U = k_2\circ s_2^U$. We can shrink $U$ if necessary so that each $U_i$ is contained in the domain of some chart for the smooth structure on $\overline{M}_2$. For each oriented chart $(U_i, \varphi_i)$, we have a corresponding chart $(U, \varphi_i\circ s_i^U)$.
Claim: The collection of such charts forms an oriented atlas for $M_2$.
To see this, let $(V, \psi_j\circ s_j^V)$ be another such chart where $U\cap V \neq \emptyset$. If $s_i^U(U)\cap s_j^V(V) \neq \emptyset$, then the transition map is just $\psi_j\circ\phi_i^{-1}$, while if $s_i^U(U)\cap s_j^V(V) = \emptyset$, the transition map is $\psi_j\circ k_2\circ\phi_i^{-1}$. In both cases, the transition map is orientation-preserving, and hence the aforementioned atlas on $M_2$ is oriented. If you would like to verify that the transition maps are as stated, we can proceed in the following way:
\begin{align*}
&\, (\psi_j\circ s_j^V)|_{U\cap V}\circ(\phi_i\circ s_i^U)^{-1}|_{\phi_i(s_i^U(U\cap V))}\\ 
=&\ \psi_j|_{s_j^V(U\cap V)}\circ s_j^V|_{U\cap V}\circ (s_i^U)^{-1}|_{s_i^U(U\cap V)} \circ\phi_i^{-1}|_{\phi_i(s_i^U(U\cap V))}\\ 
=&\ \psi_j|_{s_j^V(U\cap V)}\circ s_j^V|_{U\cap V}\circ (s_i^U|_{U\cap V})^{-1} \circ\phi_i^{-1}|_{\phi_i(s_i^U(U\cap V))}\\
=&\ \psi_j|_{s_b^{U\cap V}(U\cap V)}\circ s_b^{U\cap V}\circ (s_a^{U\cap V})^{-1}\circ\phi_i^{-1}|_{\phi_i(s_a^{U\cap V}(U\cap V))}
\end{align*}
for some $a, b \in \{1, 2\}$. If $a = b$, then
\begin{align*}
(\psi_j\circ s_j^V)|_{U\cap V}\circ(\phi_i\circ s_i^U)^{-1}|_{\phi_i(s_i^U(U\cap V))} &= \psi_j|_{s_b^{U\cap V}(U\cap V)}\circ s_b^{U\cap V}\circ (s_a^{U\cap V})^{-1}\circ\phi_i^{-1}|_{\phi_i(s_a^{U\cap V}(U\cap V))}\\ 
&= \psi_j|_{s_a^{U\cap V}(U\cap V)}\circ\phi_i^{-1}|_{\phi_i(s_a^{U\cap V}(U\cap V))}\\
&= (\psi_j\circ\phi_i^{-1})|_{\phi_i(s_a^{U\cap V}(U\cap V))}.
\end{align*}
If $a \neq b$, then $s_b^{U\cap V} = k_2|_{s_a^{U\cap V}(U\cap V)}\circ s_a^{U\cap V}$, so $s_b^{U\cap V} \circ (s_a^{U\cap V})^{-1} = k_2|_{s_a^{U\cap V}(U\cap V)}$ and therefore
\begin{align*}
(\psi_j\circ s_j^V)|_{U\cap V}\circ(\phi_i\circ s_i^U)^{-1}|_{\phi_i(s_i^U(U\cap V))} &= \psi_j|_{s_b^{U\cap V}(U\cap V)}\circ s_b^{U\cap V}\circ (s_a^{U\cap V})^{-1}\circ\phi_i^{-1}|_{\phi_i(s_a^{U\cap V}(U\cap V))}\\ 
&= \psi_j|_{s_b^{U\cap V}(U\cap V)}\circ k_2|_{s_a^{U\cap V}(U\cap V)}\circ \phi_i^{-1}|_{\phi_i(s_a^{U\cap V}(U\cap V))}\\
&= (\psi_j\circ k_2\circ\phi_i^{-1})|_{\phi_i(s_a^{U\cap V}(U\cap V))}.
\end{align*}
(The reason why this computation is harder than you might expect is because the numbering of the copies of $U$ in $\pi_2^{-1}(U)$ was completely arbitrary. The same is true of the copies of $V$ in $\pi_2^{-1}(V)$, so we can't use the indicies to determine whether $U_i$ and $V_j$ overlap. That is why we focus on the two copies of $U\cap V$ in $\pi_2^{-1}(U\cap V)$, where we can use the indices $a$ and $b$.)
In conclusion, if $k_2$ were orientation-preserving, then $M_2$ would be orientable. As $M_2$ is non-orientable, we see that $k_2$ is orientation-reversing.
Note that we didn't need compactness here.

These arguments lead to a statement which generalises both cases: if $\pi : M' \to M$ is a smooth covering with $M'$ orientable, then $M$ is orientable if and only if the deck transformations of $\pi$ are orientation-preserving.
