According to the distribution definition of the weak derivative of an $L^1_{loc}$ function, you have with $\phi\in C^\infty_c(U)$ $$ \langle \nabla_w f,\phi\rangle_{\mathscr D'(U),\mathscr D(U)}=-\int f \nabla \phi dx. $$ On the other hand you assume that the function $f$ is differentiable on $U$ with gradient $\nabla f$. Now the formulation of your question is not really meaningful: $\nabla_w f$ is defined weakly whereas $\nabla f$ has a point wise definition. So the meaning of equality or difference on a set of positive measure does not mean anything.

Yes, Hörmander's ALPDO, Chapter 22.

You are certainly right that my notations are poorly chosen: it is indeed true that $c$ can depend on $t$, via the value of $\psi$ at the critical points. Again think about multiplying everything by $e^{it}$.

@Luis Silvestre Thanks for your remarks. I certainly agree with the fact that the quoted paper by Ambrosio \& Bernard does not solve my question. The problem seems to come from the definition of the product of a measure by a measurable function, always possible in measure theory, but not in distribution theory. In particular, the definition of a weak solution for a PDE is indeed using distribution theory, with a formal integration by parts.

The $e^{it\psi}$ must be there: the value of the phase at a critical point has to be taken into account. Imagine that you multiply everything by $i$. The signature is invariantly defined as well as the square root of the determinant, which is a half-density et the critical set.

The notation $\vert \psi''(x)\vert^{1/2}$ means $$ \vert\det \psi''(x)\vert^{1/2}. $$

The $c$ in my answer is, assuming $\phi=-i\psi$, $\psi$ real-valued Morse function $$ c=\sum_{x\in supp u,d\psi(x)=0}e^{it\psi(x)}\frac{e^{i\frac{\pi}{4}sign \psi''(x)}}{\vert \psi''(x)\vert^{1/2}} u(x), $$ a coordinate-free expression. Note that $sign \psi''(x)$ stands for the signature of the quadratic form $\psi''(x)$, that is the number of positive eigenvalues minus the number of negative eigenvalues. The sum above is finite by compactness of $supp u$.

The Cauchy-Lipschitz theorem in french or in any other language is dealing with the case where $f$ is locally Lipschitz continuous with respect to $x$, i.e. satisfies an estimate of type $$ \vert f(t,x_1)-f(t,x_2)\vert\le \alpha(t) \vert x_1-x_2\vert, $$ with $\alpha \in L^1_{loc}$. In that case, local uniqueness occurs for the ODE.

$f_\epsilon$ is an entire function, thus can be approximated uniformly by polynomials on compact sets.