Relevant Background (for completeness, can be skipped)
The statement of Theorem 22.35 in Lee's Introduction to Smooth Manifolds is
Suppose $M$ is a smooth manifold, $W \subseteq T^*M$ is an open subset, $F: W \to \mathbb{R}$ is a smooth function, $S \subseteq M$ is an embedded hypersurface, and $\varphi: S \to \mathbb{R}$ is a smooth function. If the Cauchy problem (22.23)-(22.24) is noncharacteristic, then for each $p \in S$ there is a smooth solution defined on some neighborhood of $p$ in $M$.
The equations referred to in the statement are:
$$F(x, du(x)) = 0\text{ for all }x \in U\tag{22.23}$$ $$u\big\vert_S = \varphi\tag{22.24}$$
Here $U$ is some neighborhood of $p$ in $M$.
The noncharacteristic condition means that there exists a smooth 1-form $\sigma: S \to T^*M$ such that $$\sigma(x)\big\vert_{T_xS} = d\varphi(x)\text{ for all }x \in S\tag{22.25}$$ $$F(x, \sigma(x)) = 0\text{ for all }x \in S\tag{22.26}$$ and such that the following vector field along $S$ is nowhere tangent to $S$: $$A^\sigma\big\vert_x = \frac{\partial F}{\partial \xi_1}(x, \sigma(x))\frac{\partial}{\partial x^1} \bigg\vert_x+ \cdots + \frac{\partial F}{\partial \xi_n}(x, \sigma(x))\frac{\partial}{\partial x^n}\bigg\vert_x\tag{22.27}$$ where $(x^1, \ldots, x^n, \xi_1, \ldots, \xi_n)$ are natural coordinates on $T^*M$, and $\pi: T^*M \to M$ is the contangent bundle of $M$.
I can follow most of the proof, but I am stuck in one technical detail, so I will summarize the proof up to this point.
We let $\Gamma$ be the image of $\sigma$ in $T^*M$. Then $\Gamma$ is an embedded submanifold of dimension $n-1$, where $n = \dim M$. During the course of the proof, it is shown that $\Gamma$ is isotropic with respect to the canonical symplectic structure on $T^*M$.
Then we define the Hamiltonian vector field $X_F$ associated to $F$, and show that $X_F$ is nowhere tangent to $\Gamma$. We use the Hamiltonian flowout theorem to construct the flowout $\mathscr{S}$ of $\Gamma$ along $X_F$, which is an $n$-dimensional Lagrangian submanifold of $T^*M$ contained in $F^{-1}(0)$.
As shown in the proof, $\mathscr{S}$ intersects the fiber $\pi^{-1}(p)$ transversely at $(p, \sigma(p))$, and so there exist neighborhoods $U$ of $p$ in $M$ and $V$ of $(p, \sigma(p))$ in $\mathscr{S}$ such that $V$ is the image of a closed 1-form $\alpha: U \to T^*M$.
The questions
Lee says that because the image of $\sigma$ is contained in $\mathscr{S}$, it follows that $$\alpha(x) = \sigma(x)\text{ for }x \in S \cap U$$ How do we prove this? If I can show that $\alpha(S \cap U) \subseteq \Gamma$, that will be sufficient because it must lie in the image of $\sigma$, so $\alpha(x) = \sigma(y)$ for some $y$ and then $y = x$ because $x = \pi(\alpha(x)) = \pi(\sigma(y)) = y$. But I don't know how to prove that $\alpha(x) \in \Gamma$ for any $x \in S \cap U$ other than $p$. Clearly $\Gamma \cap V$ lies in the image of both $\alpha$ and $\sigma$. But perhaps for some $x \neq p$ in $S \cap U$, we could have $\alpha(x) \in V$ but $\alpha(x) \not\in \Gamma$. This would require that $\alpha(x), \sigma(x) \in \mathscr{S}$ are distinct but lie in the same fiber of $\pi$ above $x$. Intuitively, it shouldn't be possible to have $\mathscr{S}$ intersecting some fiber of $\pi$ above $S \cap U$ at more than one point, because otherwise there should be a smooth path in $\mathscr{S}$ between these two points of intersection that becomes tangent to some fiber of $\pi$ above $S$, and this should violate the noncharacteristic condition. But showing this isn't completely trivial, and I feel like I'm missing something obvious.
The neighborhood $U$ constructed above is not guaranteed to contain all of $S$, so the solution constructed in the proof does not necessarily satisfy the initial condition on $S$ outside of $U$. So what information does such a "local" solution really give about a "global" solution (one satisfying the initial condition on all of $S$)?
The fact that $\alpha(x) = \sigma(x)$ for $x\in S\cap U$ is just a reflection of the fact that $\alpha|_{S\cap U}$ and $\sigma|_{S\cap U}$ are both sections of $T^*M$ over $S\cap U$ (so that $\pi\circ\alpha(x) = x = \pi\circ\sigma(x)$ for $x\in S\cap U$), and both take their values in an open subset of $\mathscr S$ on which $\pi$ is injective (because $\alpha$ is an inverse for it).
Given the way I defined "noncharacteristic" in this context (existence of a section $\sigma$ of $T^*M$ over all of $S$ whose image lies in the zero set of $F$), you can actually produce a solution defined on an open set that contains $S$. The basic idea is to use the result of Problem 22-25, which shows that each point of $S$ has a neighborhood $U$ on which there is a unique solution satisfying the additional condition $du|_x = \sigma(x)$ for all $x\in S\cap U$. Using this uniqueness, you can piece together the local solutions to create one defined on an open set containing all of $S$. But it's a little tricky, because you might have to shrink the local neighborhoods to ensure that the solutions obtained by flowing out from different points of $S$ don't overlap each other. (I mentioned an analogous difficulty in the discussion of nonuniqueness of solutions to quasilinear equations on pp. 244-245.) The technique for identifying a suitable open neighborhood of $S$ will be similar to the technique for proving the existence of a tubular neighborhood (Thm. 6.24).
I should mention that it's possible to give a more general definition of "noncharacteristic," for which there might not be a solution defined on an entire neighborhood of $S$. One could weaken the definition of noncharacteristic by requiring only that for each $p\in S$, there exists a neighborhood $W$ of $p$ in $S$ and a section $\sigma\colon W\to T^*M$ taking its values in $F^{-1}(0)$. In that case, it might not be possible to find a $\sigma$ defined on all of $S$. For example, let $E$ denote the total space of the Möbius bundle defined in Example 10.3 -- it's the quotient of $\mathbb R^2$ by the $\mathbb Z$-action defined by $n\centerdot (x,y) = (x+n, (-1)^n y)$. It inherits a Riemannian metric from the flat metric on $\mathbb R^2$. Define $F\colon T^*E\to \mathbb R$ by $F(x,\xi) = |\xi|^2-1$, and let $S\subset E$ be the image of the zero section: $S = \{[0,y]: y\in\mathbb R\}$. Then there is no function $u$ defined on a neighborhood of $S$ that satisfies $F(x,du(x)) = 0$ and $u|_S=0$. An easy way to see this is to note that the gradient of $u$ would have to be nonzero along $S$ and tangent to the fibers of $E$, so under the canonical identification $T_xE \cong E_x$ for $x\in S$, it would yield a nonvanishing section of the bundle $E$, which is impossible.