More generally, let $k$ be a field of characteristic $≠2$. For the analogous question over $k$, a necessary and sufficient condition is that the doubling map $[2]_E:E\to E$ is not surjective on $k$-points.

Of course, if $k$ is finite, this is turn is equivalent to "$\#E(k)$ is even".

Indeed, given a polynomial $g$ as in the question, consider the divisor $D$ in $E$ given by zeros of $g$. Over $\overline{k}$, $D$ must be a translate of $E[2]=\ker[2]_E$ by some point $P\in E(\overline{k})$, and the image of $D$ by $[2]_E$ is $Q:=2P$ which must be $k$-rational because $D$ is. Then it is immediately checked that the double cover $\pi:E\to\mathbb{P}^1_k$ associated to $g$ is the quotient of $E$ by the involution $x\mapsto Q-x$.

Conversely, any $Q\in E(k)$ gives rise to an involution $\sigma:x\mapsto Q-x$. The quotient $E/\langle\sigma\rangle$ is smooth of genus zero and has a rational point because $E$ does, hence $E/\langle\sigma\rangle\cong\mathbb{P}^1_k$. The branch divisor of $E\to E/\langle\sigma\rangle$ on $E$ is $[2]_E^{-1}(Q)$, so the desired condition is satisfied iff $Q$ is not divisible by $2$ on $E(k)$.

More generally, let $k$ be a field of characteristic $≠2$. For the analogous question over $k$, a necessary and sufficient condition is that the doubling map $[2]_E:E\to E$ is not surjective on $k$-points.

Of course, if $k$ is finite, this in turn is equivalent to "$\#E(k)$ is even".

Indeed, given a polynomial $g$ as in the question, consider the divisor $D$ in $E$ given by zeros of $g$. Over $\overline{k}$, $D$ must be a translate of $E[2]=\ker[2]_E$ by some point $P\in E(\overline{k})$, and the image of $D$ by $[2]_E$ is $Q:=2P$ which must be $k$-rational because $D$ is. Then it is immediately checked that the double cover $\pi:E\to\mathbb{P}^1_k$ associated to $g$ is the quotient of $E$ by the involution $x\mapsto Q-x$.

Conversely, any $Q\in E(k)$ gives rise to an involution $\sigma:x\mapsto Q-x$. The quotient $E/\langle\sigma\rangle$ is smooth of genus zero and has a rational point because $E$ does, hence $E/\langle\sigma\rangle\cong\mathbb{P}^1_k$. The branch divisor of $E\to E/\langle\sigma\rangle$ on $E$ is $[2]_E^{-1}(Q)$, so the desired condition is satisfied iff $Q$ is not divisible by $2$ on $E(k)$.

More generally, let $k$ be a field of characteristic $≠2$ such that every smooth $k$-conic has a rational point (e.g. $\mathrm{Br}(k)=0$). For the analogous question over $k$, a necessary and sufficient condition is that the doubling map $[2]_E:E\to E$ is not surjective on $k$-points. Of course, if $k$ is finite, this is turn is equivalent to "$\#E(k)$ is even".

Indeed, given a polynomial $g$ as in the question, consider the divisor $D$ in $E$ given by zeros of $g$. Over $\overline{k}$, $D$ must be a translate of $E[2]=\ker[2]_E$ by some point $P\in E(\overline{k})$, and the image of $D$ by $[2]_E$ is $Q:=2P$ which must be $k$-rational because $D$ is. Then it is immediately checked that the double cover $\pi:E\to\mathbb{P}^1_k$ associated to $g$ is the quotient of $E$ by the involution $x\mapsto Q-x$.

Conversely, any $Q\in E(k)$ gives rise to an involution $\sigma:x\mapsto Q-x$ with quotient$E/\langle\sigma\rangle$ of genus zero, hence isomorphic to $\mathbb{P}^1_k$ by our assumption on $k$. The branch divisor on $E$ is $[2]_E^{-1}(Q)$, so the desired condition is satisfied iff $Q$ is not divisible by $2$ on $E(k)$.

More generally, let $k$ be a field of characteristic $≠2$. For the analogous question over $k$, a necessary and sufficient condition is that the doubling map $[2]_E:E\to E$ is not surjective on $k$-points. 

Of course, if $k$ is finite, this is turn is equivalent to "$\#E(k)$ is even".

Indeed, given a polynomial $g$ as in the question, consider the divisor $D$ in $E$ given by zeros of $g$. Over $\overline{k}$, $D$ must be a translate of $E[2]=\ker[2]_E$ by some point $P\in E(\overline{k})$, and the image of $D$ by $[2]_E$ is $Q:=2P$ which must be $k$-rational because $D$ is. Then it is immediately checked that the double cover $\pi:E\to\mathbb{P}^1_k$ associated to $g$ is the quotient of $E$ by the involution $x\mapsto Q-x$.

Conversely, any $Q\in E(k)$ gives rise to an involution $\sigma:x\mapsto Q-x$. The quotient  $E/\langle\sigma\rangle$ is smooth of genus zero and has a rational point because $E$ does, hence $E/\langle\sigma\rangle\cong\mathbb{P}^1_k$. The branch divisor of $E\to E/\langle\sigma\rangle$ on $E$ is $[2]_E^{-1}(Q)$, so the desired condition is satisfied iff $Q$ is not divisible by $2$ on $E(k)$.