Suppose that $M$ is an $n-$ dimensional smooth manifold and if there exists an atlas consisting of only two charts $(U,x)$ and $(V,y)$. If $U\cap V$ is connected, then $M$ is orientable.
To prove this, I must show that the transition map $x\circ y^{-1}: y(U\cap V)\to x(U\cap V)$ has a positive Jacobian determinant. That is, if $x=(x_1,...,x_n), y=(y_1,...,y_n)$, then the determinant of the matrix $(\partial x_i/\partial{y_j})_{i,j}$ is positive. But I am not sure how to show that.
Since $U\cap V$ is connected and $\det$ is continuous, it suffices to show that the determinant is positive only at a point $p\in U\cap V$.
Consider the standard orientation given by $dr_1\wedge \cdots\wedge dr_n$ on $\mathbb R^n$. Let's pull these back to $U$ and $V$:
$\begin{align}x^\ast (dr_1\wedge \cdots\wedge dr_n)&=d(r_1\circ x)\wedge\cdots\wedge d(r_n\circ x)\\ &=dx_1\wedge\cdots\wedge dx_n\\ &=\det\left((\partial x_i/\partial{y_j})_{i,j}\right)dy_1\wedge\cdots\wedge dy_n\\ &=\det\left((\partial x_i/\partial{y_j})_{i,j}\right)y^\ast (dr_1\wedge \cdots\wedge dr_n)\end{align}$
If $x,y$ are orientation preserving then the determinant is $>0$ and we are done.
If $x,y$ are both orientation reversing, then also the determinant is $>0$ and we are done.
But what if $x$ is orientation preserving and $y$ is not? In this situation, the determinant could be negative. How to contradict this?
You can’t get a contradiction. Note you’re asked to prove orientability, not that the specific $\{(U,x),(V,y)\}$ is an oriented atlas (which is a false statement). So, you should instead replace one of the coordinate functions, say $y^1$ with $-y^1$ if you get a negative determinant; then the new pair will have transition with positive determinant.
Also, your ‘proof’ doesn’t really make sense. You can’t speak of $x,y$ being orientation preserving/reversing unless you have an orientation on $M$ already. So, you should rephrase your argument: