For any linear operator $A$, the adjoint $A^*$ is defined as a linear operator that satisfies
$$\langle v, Au\rangle = \langle A^*v, u\rangle$$
Moreover, one has that $(A^{*})^{*} = A$ (proof here). When things are nonlinear, this seems to result in weird conclusions. Let $F$ be an operator that is not linear.
Can one define the adjoint of $F$ and if yes how? If one goes along with the usual definition, then we see that
$\begin{align} \langle x, F(y+z)\rangle &= \langle F^*x, y+z\rangle \\ &= \langle F^*x, y\rangle +\langle F^*x, z\rangle \\ &= \langle x, F(y) + F(z)\rangle \end{align}$
I'm not sure if this explicitly contradicts the assumption that $F$ is nonlinear but it seems to? I did assume that $(F^*)^* = F$ but the linked proof above doesn't seem to require linearity so this seems okay.
In general, is it possible to define the adjoint in the sense of $\langle v, Fu\rangle = \langle F^*v, u\rangle$ for nonlinear $F$?
Let $V$ be a vector space with scalar product $\langle,\rangle$. Let $F,F^*\colon V\to V$ be maps with $\langle F^*u,v\rangle=\langle u,Fv\rangle$ for all $u,v$. Then $$\langle u,F\lambda v\rangle=\langle F^*u,\lambda v\rangle=\lambda\langle F^*u,v\rangle=\lambda\langle u,Fv\rangle=\langle u,\lambda Fv\rangle $$ It follows that $F\lambda v-\lambda Fv$ is perpendicular to all $u\in V$, i.e., $F\lambda v=\lambda Fv$ for all $v\in V$ and all $\lambda\in K$.
Additionally, your own computation shows (indeed without assumingn that $F^{**}=F$!) that also $Fy+Fz-F(y+z)$ is perpendicular to all $x\in V$, i.e., $F$ is also additive.
We conclude that $F$ is linear. By essentially the same argument, $F^*$ is linear.