I am trying to solve exercise 3.16 of do Carmo's Differential Forms and Applications. The problem is as follows:
Let $M$ be a connected differentiable manifold. For each $p\in M$, denote by $\mathcal{O}_p$ the quotient space of the set of all bases of $T_pM$ under the following equivalence relation: tow bases are equivalent if they are related by a matrix with positive determinant. Clearly $\mathcal{O}_p$ has two elements, and each element $O_p$ of $\mathcal{O}_p$ is called an orientation at p. Now let $$\tilde{M}=\{(p,O_p)\mid p\in M,O_p\in\mathcal{O}\},$$ and let $f_\alpha:U_\alpha\rightarrow M$ be a parametrization of $M$ with $p\in f_\alpha(U_\alpha)$. Define $\tilde{f}_\alpha:U_\alpha\rightarrow \tilde{M}$ by $\tilde{f}_\alpha(x_1,\ldots,x_n)=(f_\alpha(x_1,\ldots,x_n),[\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_1}])$ where $(x_1,\ldots,x_n)\in U_\alpha$, and $[\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_1}]$ denotes the element of $\mathcal{O}_p$ determined by this basis. Show that if $\{(U_\alpha,f_\alpha)\}$ is a differentiable structure in $M$, then $\{(\tilde{U}_\alpha,\tilde{f}_\alpha)\}$ is a differentiable structure in $\tilde{M}$ which is orientable (even if $M$ is not).
What I am having trouble understanding is how $\{(U_\alpha,f_\alpha)\}$ is a differentiable structure on $\tilde{M}$. In particular, I can't see why $\bigcup_\alpha\tilde{f}_\alpha(U_\alpha)=\tilde{M}$. The basis $\{\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_1}\}$ should be determined at each $p$ by $f_\alpha$ so that given some $p\in U_\alpha$, $\tilde{f}_\alpha(p)$ only maps $p$ to one component of $\tilde{M}$. Thus, unless $p$ happens to be included in some $U_\beta$ with $f_\beta$ that provides an oppositely oriented basis for $T_pM$, I see no reason why $(f_\alpha(p),+)$ and $(f_\alpha(p),-)$ (where by $+$ and $-$ I denote the two elements of $\mathcal{O}_p$) should appear in the image of the coordinate charts.