Forget about $x,t$. Consider a $C^1$ mapping $\phi:(\zeta,\eta)\mapsto (u,v)$. Locally if $|d\phi|\neq 0$ we can invert it. Let the inverse be $\psi: (u,v)\mapsto (\zeta,\eta)$, so $\psi\circ\phi(\zeta,\eta) = (\zeta,\eta)$ and $\phi\circ\psi(u,v) = (u,v)$.

An elementary computation shows that the Jacobian matrices of $\phi$ and $\psi$ are inverses. That is, evaluated at a fixed $u,v,\zeta,\eta$ where $(u,v) = \phi(\zeta,\eta)$,

$$ \begin{pmatrix} \partial_1\psi^1 & \partial_1\psi^2 \newline \partial_2\psi^1 & \partial_2\psi^2\end{pmatrix}^{-1} = \begin{pmatrix} \partial_1\phi^1 & \partial_1\phi^2 \newline \partial_2\phi^1 & \partial_2 \phi^2\end{pmatrix} $$

For disambiguation: write the function $\phi = (U,V)(\zeta,\eta)$ and the function $\psi = (Z,N)(u,v)$. The above implies

$$ \begin{pmatrix} U_\zeta & V_\zeta \newline U_\eta & V_\eta\end{pmatrix} = \frac{1}{Z_u N_v - Z_v N_u} \begin{pmatrix} N_v & - N_u \newline -Z_v & Z_u\end{pmatrix} $$

Your system (1) gives $U_\eta = V_\zeta$ which from the matrix inequality immediately implies that $Z_v = N_u$, which is the first equation of system (2).

Using that $Z_uN_v - Z_vN_u = |d\psi| \neq 0$, the equality of the matrix components also gives you that the second equation of (2) is obtained from the second equation of (1) by the replacement $U_\zeta \to N_v$, $V_\zeta \to -N_u$, $U_\eta \to -Z_v$ and $V_\eta \to Z_u$.

**Notice that this works because (a) we have a 2-by-2 system and (b) the system (1) is ***quasilinear*. If it were genuinely nonlinear, we cannot "factor" out from the equation the common factor of $|d\psi|$ to end up with a purely linear equation.

And yes, this is an example of the hodographic transformation. In general the same procedure works for any quasilinear system of first order partial differential equations with 2 dependent and 2 independent variables.