A functional equation on $\mathbb{Z}/p\mathbb{Z}$

Claim. The one and only function $f$ that satisfies the conditions stated above is the identity function.

Proof. Put $N=p-1$. Since $2$ is a primitive root mod $p$, every $x\in\mathbb F_p^\times$ can be written uniquely as $x=2^n$ with $n\in\mathbb Z/N\mathbb Z$. Define $$ g:\mathbb Z/N\mathbb Z\to\mathbb Z/N\mathbb Z $$ by $f(2^n)=2^{g(n)}$. Because $f$ is a bijection on $\mathbb F_p^\times$, the map $g$ is a permutation of $\mathbb Z/N\mathbb Z$.

Plugging $x=2^n$ into $f(2x)=f(x)+f(f(x))$ gives, for all $n\in\mathbb Z/N\mathbb Z$,

$$ 2^{g(n+1)} = 2^{g(n)} + 2^{g(g(n))}. \tag{1} $$

If $f$ has a fixed point, then so does $g$: $$ f(2^n)=2^n \iff g(n)=n. $$ Moreover, if $g(n_0)=n_0$ for some $n_0$, then $(1)$ evaluated at $n=n_0$ gives $$ 2^{g(n_0+1)}=2^{n_0}+2^{n_0}=2^{n_0+1}, $$ so $g(n_0+1)=n_0+1$, and iterating yields $g(n)=n$ for all $n$, hence $f(x)=x$ for all $x$. So, if $g$ has a fixed point, then so does $f$.

Define vectors $u,v\in\mathbb F_p^N$ (indexed by $\mathbb Z/N\mathbb Z$) by $u_n:=2^n$ and $v_n:=2^{g(n)}$. Let $S$ be the cyclic shift operator $(Sw)_n:=w_{n+1}$, and let $P_g$ be the permutation operator $(P_gw)_n:=w_{g(n)}$. We can easily see that $(1)$ is exactly $$ (Sv)_n = v_n + (P_g v)_n $$ for all $n$, i.e. $$(S-I)v = P_g v.$$

Substitute $v=P_g u$: $$ (S-I)P_g u = P_g^2 u $$

Left-multiply by $P_g^{-1}$ and set $C:=P_g^{-1}SP_g$: $$ (C-I)u = P_g u $$

If we look at the indices (which is only natural, because these operators act, after all, on the indices of some vectors), $S$ corresponds to the $N$-cycle $s(n)=n+1$, hence $C$ corresponds to the conjugate permutation $c := g^{-1}\circ s\circ g$, i.e. $c(n)=g^{-1}(g(n)+1)$. The relevant formulation of this identity being $$ g(c(n))=g(n)+1 \tag{2} $$ for all $n$.

Now, the equation $$ (C-I)u=P_g u $$ may be written as: $$ u_{c(n)}-u_n = u_{g(n)}. $$ Since $u_n=2^n$, this becomes $$ 2^{c(n)} - 2^n = 2^{g(n)} \tag{3} $$ for all $n$.

Fix $n_0\in\mathbb Z/N\mathbb Z$ and define the $c$-orbit $n_k:=c^k(n_0)$ for $k=0,1,\dots,N-1$. Because $c$ is an $N$-cycle, the residues $n_0,n_1,\dots,n_{N-1}$ are all distinct and run through all of $\mathbb Z/N\mathbb Z$. From $(2)$, applied repeatedly along the orbit, we get

$$ g(n_k)=g(n_0)+k \pmod N \tag{4} $$ for all $k$.

Look at $(3)$ and apply $(4)$ with $n=n_k$: $$ 2^{n_{k+1}} - 2^{n_k} = 2^{g(n_k)} = 2^{g(n_0)+k} \tag{5} $$ for all $k$.

It is trivial to see that if we sum $(5)$ from $k=0$ to $t-1$: $$ 2^{n_t}-2^{n_0} = \sum_{k=0}^{t-1}2^{g(n_0)+k} = 2^{g(n_0)}(2^t-1). $$

Thus $$ 2^{n_t} = A\cdot 2^t + B, \tag{6} $$ where $A:=2^{g(n_0)}\neq 0$ and $B:=2^{n_0}-2^{g(n_0)}$.

Now we use the fact that $2$ is a primitive root mod $p$ once more: as $t$ runs through $0,1,\dots,N-1$, the values $2^t$ run through all of $\mathbb F_p^\times$; and since $n_t$ runs through all residues mod $N$, the values $2^{n_t}$ also run through all of $\mathbb F_p^\times$. So the affine map $\varphi(x)=Ax+B$ satisfies $\varphi(\mathbb F_p^\times)=\mathbb F_p^\times$.

If $B\neq 0$, then $\varphi(x)=0$ has the solution $x=-B/A$, and since $A\neq 0$ this solution is valid and nonzero, i.e. $x\in\mathbb F_p^\times$. That would force $0\in\varphi(\mathbb F_p^\times)$, which is contradictory. Hence $B=0$ in $(6)$, so $2^{n_0}=2^{g(n_0)}$, hence $g(n_0)=n_0$.

Thus $g$ has a fixed point, hence $g=\mathrm{id}$, hence $f=\mathrm{id}$ on $\mathbb F_p^\times$. $$\tag*{$\Box$}$$

Remark: Curiously, we did not use the fact that $f$ is an odd function, but, as it has been discussed in the comments, bijectivity was necessary for our proof.