Let $F:\Bbb R \to \Bbb R^n$, and consider the PDE $$u_t +\nabla \cdot F(u)=0$$for a function $u:\Bbb R^{n+1}\to \Bbb R$. Then the claim is that $u=g(x-tF'(u))$ defines an implicit solution to the PDE.
I could justify this without really understanding what was going on (and you can say that my question is actually about Calculus instead of PDE, I'll have to agree with you).
1) my first issue is with the notation $\nabla\cdot F(u)$. It stands to reason that even though $F$ is not a vector field per se, one freezes $t$ and gets the vector field $x\mapsto F(u(x,t))$. Then take the divergence of that, which turns out to be $\langle F'(u),\nabla_xu\rangle$, where $\nabla_xu$ is the gradient of $u$ taken only relative to space variables.
2) if $u=g(x-tF'(u))$, then I just computed $u_t =\langle -F'(u),\nabla g\rangle$ and $\nabla g =\nabla_xu$. Plug that in the PDE and immediately get the zero we want.
But I'm not comfortable differentiating $u=g(x-tF'(u))$. For starters we have this abuse of notation, meaning $$u(x,t) = g(x-tF'(u(x,t)))$$instead, so why the chain rule doesn't give further derivatives of $u$ in the right side?
Your doubts are justified. The computation in 2) is only valid for cases where $u\mapsto F'(u)$ is constant. On the one hand, the chain rule gives $$ u_t = -(F'(u) + t F'(u)_t)\cdot \nabla g(x - t F'(u)) . $$ On the other hand, the chain rule gives $$ \nabla u = (I_n - t\nabla F'(u))\cdot \nabla g(x - t F'(u)) $$ i.e. $$ F'(u)\cdot \nabla u = (F'(u) - tF'(u)\cdot\nabla F'(u))\cdot \nabla g(x - t F'(u)) . $$ The sum of both equations gives \begin{aligned} u_t + F'(u)\cdot \nabla u &= -t\, (F'(u)_t + F'(u)\cdot\nabla F'(u))\cdot \nabla g(x - t F'(u)) \\ &= -t\, F''(u)\cdot (u_t + F'(u)\cdot\nabla u)\cdot \nabla g(x - t F'(u)) \\ &= 0 \, , \end{aligned} which ends the proof (there might be some missing transposition symbols...).
To learn more about it, I'd recommend to work out the scalar case $n=1$ first. For the computation of the gradients and divergences, I'd recommend to introduce orthonormal Cartesian coordinates and to work componentwise (e.g., with Einstein notation).