Every connected orientable smooth manifold has exactly two orientations, Lee Proposition 15.9

I think that is correct. One can also do that by considering orientation forms as follows.

Let $(M,\mathcal{O})$ be the connected smooth oriented manifold. By hypothesis, there exists a nonvanishing $n$-form $\omega$ on $M$ such that $\omega$ is positively oriented at each point. Any other choice of orientation $\widetilde{\mathcal{O}}$ for $M$ will induce a nonvanishing $n$-form $\widetilde{\omega}$ on $M$. Since $\omega = f \widetilde{\omega}$ for some smooth function $f : M \to \mathbb{R}$ such that $f(p) \neq 0$ for all $p \in M$, then by the connectedness of $M$, the image $f(M) \subseteq \mathbb{R}$ is a connected subset which does not contain $0$. I.e., the function $f$ either always positive or negative on $M$. If $f$ is positive, then $\omega_p$ and $\widetilde{\omega}_p$ determine the same orientation at $T_pM$ for each $p \in M$. Hence $\widetilde{\mathcal{O}} = \mathcal{O}$. So lets assume that $f$ is negative. By the similar argument $\omega_p$ and $\widetilde{\omega}_p$ determine different orientation at $T_pM$ for each $p \in M$. So $\mathcal{O}$ is different than $\widetilde{\mathcal{O}}$. By similar argument as above, any other orientation $\mathcal{O}'$ for $M$ will induce a non-vanishing $n$-form $\omega'$ such that $\omega' = g \omega$ for either $g : M \to \mathbb{R}$ is positive or negative function. If $g$ is positive, then $\mathcal{O}' = \mathcal{O}$. If $g$ is negative then $\mathcal{O}'$ different than $\mathcal{O}$ and then the product $gf$ is a positive function. Hence above relation $\omega' =g \omega= gf \widetilde{\omega}$ implies that $\mathcal{O} = \widetilde{\mathcal{O}}$. This proves that there are exactly two orientation on $M$.

Suppose $\mathcal{O}_1$ and $\mathcal{O}_2$ are orientation for $M$ and $\omega_1$ and $\omega_2$ are the orientation forms for $\mathcal{O}_1$ and $\mathcal{O}_2$ respectively. Let $\omega_1 = f \omega_2$. If they agree at a point $p \in M$, then the orientation forms is positive multiples of each other at $p$. This implies that $f$ is a positive function. Hence $\omega_1$ and $\omega_2$ determines the same orientation for each point in $M$, i.e., $\mathcal{O}_1 = \mathcal{O}_2$.