Let $\omega$ be an $n$-form on an $n$-manifold $M$, and $F: M\to M$ be a diffeomorphism. Here for simplicity let $\phi:U\to S$ and $\psi:V\to T$ be two charts of $M$ such that $F(U) = V$.
I believe we have the definition $\int_U \omega = \int_S (\phi^{-1})^* \omega$, and similar for $\int_V \omega$.
I believe we have $\int_U F^* \omega = \pm \int_V \omega$, positive if $F$ is orientation-preserving, and negative otherwise. I am attempting to understand the this equality.
Let $x^k$ be the local coordinates on $U$, and $y^k$ be that on $V$. Write $\omega = \omega_{12...n} dx^1 \wedge dx^2 \wedge \cdots \wedge dx^n$ on $U$ and $\omega = \omega_{12...n}' dy^1 \wedge dy^2 \wedge \cdots \wedge dy^n$ on $V$.
Also let $f: S\to T$ be the local representation of $F$.
To make things absolutely clear, I am also going to write $s^k$ for coordinates on $S$, and $t^k$ for that on $T$.
Now
$\begin{equation}
\begin{aligned}
\int_U F^* \omega &= \int_S (\phi^{-1})^* F^* \omega \\
&= \int_S f^* (\psi^{-1})^* \omega \\
&= \int_S f^* (\omega_{12...n}'\circ \psi^{-1} dt^1 \wedge dt^2 \wedge \cdots \wedge dt^n) \\
&= \int_S (\omega_{12...n}'\circ \psi^{-1} \circ f) \det(df) \cdot ds^1 \wedge ds^2 \wedge \cdots \wedge ds^n \\
&= \int_S (\omega_{12...n}'\circ \psi^{-1} \circ f(s)) \det(df)(s) \cdot ds^1 ds^2 \dots ds^n
\end{aligned}
\end{equation}$
Then by multivariable calculus the above becomes
$\text{sign}\det(df) \int_T \omega_{12...n}'\circ \psi^{-1} (t) \cdot dt^1 dt^2 \dots dt^n
= \text{sign}\det(df) \int_T (\psi^{-1})^* \omega
= \text{sign}\det(df) \int_V \omega$.
If everything so far is correct, then $\text{sign}\det(df)$ should correspond to whether $F$ is orientation-preserving.
But what if I change the chart $\psi$ to $\psi_0: V\to T_0$, where $\psi_0 (v)$ switches the first two coordinates of $\psi(v)$? Then we have a new set of coordinates $z^k$ on $V$ and $u^k$ on $T_0$, new expression $\omega = \omega_{12...n}'' dz^1 \wedge dz^2 \wedge \cdots \wedge dz^n$ on $V$, and a new local representation $f_0: S\to T_0$ of $F$.
We do the exact same computation as before. From what I can see nothing seems to break, and we still get $\int_U F^* \omega = \text{sign}\det(df_0) \int_V \omega$.
But $df_0$ switches the first two columns of $df$, so $\det(df_0)$ and $\det(df)$ must have different signs.
This seems to be not great. For the life of me I couldn't figure out where exactly things went wrong.