Rigiorous justification for method of characteristics applied to quasilinear PDEs 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 $C$.
 A: 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 $S$.
