We can fix the extension part of the solution as follows. Put $U=\left( -\tfrac 13, \tfrac 23\right)$.
We claim that for each $x_1,\dots, x_k\in U$ with $x_1+\dots+x_k=0$ we have $g(x_1)+\dots+g(x_k)=0$. Let’s prove this claim by induction with respect to $k$. For $k\le 3$ the claim is given. Assume that the claim is proved for each $k\le n\ge 3$. Let $x_1,\dots, x_{n+1}\in U$ with $x_1+\dots+x_{n+1}=0$. Without loss of generality we can assume that $x_1\le 0\le x_2$, so $x_1+x_2\in [x_1, x_2]\subset U$. By the inductive hypothesis we have $$g(x_1+x_2)+g(x_3)+\dots+g(x_n)=0,$$ so it remains to prove that $g(x_1+x_2)=g(x_1)+g(x_2)$. It is easy to see that $-\tfrac{x_1+x_2}2\in U$, so $$g(x_1+x_2)+2g\left(-\tfrac{x_1+x_2}2\right)=0.$$ Similarly we have $$g(x_1)+2g\left(-\tfrac{x_1}2\right)=0\mbox{ and }g(x_2)+2g\left(-\tfrac{x_2}2\right)=0.$$ Moreover, $$g\left(-\tfrac{x_1+x_2}2\right)+ g\left(\tfrac{x_1}2\right)+ g\left(\tfrac{x_2}2\right)=0,$$
$$g\left(\tfrac{x_1}2\right)+ g\left(-\tfrac{x_1}2\right)=0,\mbox{ and } g\left(\tfrac{x_2}2\right)+ g\left(-\tfrac{x_1}2\right)=0.$$ It follows
$$g(x_1+x_2)=$$ $$-2g\left(-\frac{x_1+x_2}2\right)=2 g\left(\frac{x_1}2\right)+2g\left(\frac{x_2}2\right)=-2 g\left(-\frac{x_1}2\right)-2g\left(-\frac{2}2\right)=$$ $$g(x_1)+g(x_2).$$
Let $x\in\Bbb R$ be any number, $x=x_1+\dots+x_n$ and $x=x’_1+\dots+x’_m$ be two representations of $x$ with $x_1,\dots, x_n, x’_1,\dots, x’_m\in U$. Then $\pm \tfrac {x_i}2$ and $\pm \tfrac {x’_j}2$ belong to $U$ for each $i$ and $j$. By the claim we have
$$g(x_1)+\dots+g(x_n)=$$
$$-2\left(g\left(-\frac{x_1}2\right)+\dots+ g\left(-\frac{x_n}2\right) \right)=$$
$$2\left(g\left(\frac{x’_1}2\right)+\dots+ g\left(\frac{x’_m}2\right) \right)=$$
$$-2\left(g\left(-\frac{x’_1}2\right)+\dots+ g\left(-\frac{x’_m}2\right) \right)=$$
$$g(x’_1)+\dots+g(x’_n).$$
Put $h(x)=g(x_1)+\dots+ g(x_n)$. The definition of $h(x)$ implies that $h$ is additive and an extension of $g$. The uniqueness of such an $h$ follows from its additivity and the claim, but I guess it is not needed for the solution, since existence of any additive extension of $g$ on $\Bbb R$ implies $g(x)=kx$ for some $k\in\left[-\tfrac 12,1\right]$.