Let \begin{equation} a(t,x,u) \frac{\partial u}{ \partial t} + b(t,x,u) \frac{\partial u}{ \partial x} - c(t,x,u) = 0, \\ u(0,x) = \varphi(x), \end{equation} be a quasilinear, $2$-dimensional first oder PDE with $a,b,c \in \mathcal{C}^{1}(\mathbb{R}^3, \mathbb{R})$ and $\varphi \in \mathcal{C}^{1}(\mathbb{R}, \mathbb{R})$.
Now consider the solution surface of the PDE, given by \begin{equation} C = \{ (t,x,u(t,x)) \mid t,x \in \mathbb{R} \}. \end{equation} I understand, that for geometric reasons, the vector field \begin{equation} V : \mathbb{R}^3 \to \mathbb{R}^3, \quad V(t,x,u) = (a(t,x,u), b(t,x,u), c(t,x,u)), \end{equation} must be tangential to $C$ at every point $c \in C$. Therefore, it seems not too crazy to assume, that solving the ODE \begin{equation} \gamma^{\prime} = V(\gamma), \quad \gamma(0,0,0) = (0,x,\varphi(x)) \in C, \end{equation} should deliver a solution trajectory $\gamma$ which lies fully in $C$. This is the reasoning behind the method of characteristics, which tried to express $C$ as a union of disjoint trajectories $\gamma$ with different initial starting points.
But why is this true? The mere fact, that $\gamma$ is tangential to $C$ at its initial value seems insufficient to conclude, that $\gamma$ always stays in $C$.
One idea is to "do everything in charts" and then go back to the surface.
Given a regular surface $S \subset \mathbb{R}^3$ we can parametrize $S$ (locally, say near $p \in S$) by $\phi : U \to \mathbb{R}^3$, where $U \subset \mathbb{R}^2$ is open and $\phi$ is a diffeomorphism. In your above example, $\phi(t, x) = (t, x, u(t, x))$. Let $u, v$ denote the coordinates in the parameter space $U$, so that $\phi_u, \phi_v$ (subscripts denote partials) span the tangent space to $S$. We can then write the vector field on $S$ as $V(q) = V^u(q)\phi_u(u, v) + V^v\phi_v(u, v)$ where $q = \phi(u, v) \in S$. This pulls back to the vector field $\tilde{V}(u, v) = ({V}^u(u, v), {V}^v(u, v))$. Note $V = d\phi(\tilde{V})$.
We then solve the ODE $\eta'(t) = \tilde{V}(\eta(t))$ in $\mathbb{R}^2$. Define $\gamma = \phi \circ \eta$; then $$ \gamma'(t) = d\phi_{\eta(t)}(\eta'(t)) = d\phi_{\eta(t)}(\tilde{V}(\eta(t))) = V(\phi(\eta(t))) = V(\gamma(t)). $$ Thus, $\gamma$ solves the original ODE and by construction lies entirely within $S$. If you like, uniqueness of solutions to ODEs in $\mathbb{R}^3$ shows that this is the solution to your original ODE, and thus that said solution must, in fact, lie in the surface $S$.