= Solution
All geometric <2-torsion> is rational: its points are $O,(\alpha,0),(\beta,0),(\gamma,0)$. We will construct the required finite quotient directly, without assuming any theorem about finite generation or <two-descent on an elliptic curve>.
Choose $Q\in E(\overline K)$ with $2Q=P$ and define the <halving cocycle with rational two-torsion>
$$
c_P(\sigma)=\sigma Q-Q\in E[2],\qquad \sigma\in G_K.
$$
Such a half exists because the nonconstant multiplication morphism $[2]$ on a smooth projective <elliptic curve> is surjective over an <algebraic closure>. Since $E[2]$ is rational, $c_P(\sigma\tau)=c_P(\sigma)+c_P(\tau)$. A different half is $Q+T$ with rational $T\in E[2]$ and gives the same cocycle. Replacing $P$ by $P+2R$, with $R\in E(K)$, allows the half $Q+R$ and again changes nothing. Adding chosen halves shows additivity in $P$. If the cocycle is zero, the half is fixed by the <absolute Galois group> and lies in $E(K)$, so $P\in2E(K)$. This proves, rather than assumes, an injection
$$
E(K)/2E(K)\hookrightarrow\operatorname{Hom}_{\mathrm{cont}}(G_K,E[2]).
$$
The fixed field of $\ker c_P$ is exactly $K(Q)$. Its <Galois group> is the image of $c_P$, so its degree is at most four. This also follows from the permitted biquadratic description.
Let $S$ be the finite set of primes dividing $2(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)$. At a place $v\notin S$ the displayed integral cubic has unit <elliptic-curve discriminant>, hence <good reduction of an elliptic curve>, and the residue characteristic is odd. Choose a half $\overline Q$ of the reduction of $P$ over the algebraic closure of the <residue field>. Its coordinates belong to a finite residue extension; let $L_v/K_v$ be the corresponding finite <unramified extension>. Smoothness and the <Hensel lemma> lift $\overline Q$ to some $Q_0\in E(L_v)$. The difference $P-2Q_0$ reduces to zero and belongs to the <formal kernel of a minimal Weierstrass equation>. In its integral <formal group law> the doubling series is $2t+O(t^2)$, with unit linear coefficient. The <prime-to-residue-characteristic multiplication on a formal group> gives a unique $R$ in this kernel such that $2R=P-2Q_0$. Thus $Q_0+R$ is a half of $P$ defined over an unramified local extension. Every other half differs by rational <2-torsion>, so it too is unramified. We have proved the <unramified halving fields for a split cubic> assertion at every $v\notin S$.
There are only finitely many <bounded-degree extensions with restricted ramification> of $K$ of degree at most four. This is the permitted number-field finiteness consequence of the <Hermite–Minkowski theorem>: bounded local degrees bound the discriminant exponents at the finitely many allowed ramified primes, so absolute discriminants are bounded. Each possible halving field $L$ is Galois, and there are only finitely many homomorphisms $\operatorname{Gal}(L/K)\to E[2]$. The injection constructed above therefore has finite image. \b[Consequently $E(K)/2E(K)$ is finite.] No elliptic-curve finiteness or descent theorem was used in this argument.
For $K=\mathbb Q$, write $s=\#S$. Each coordinate of $c_P$ in a basis of $E[2]\cong(\mathbb Z/2\mathbb Z)^2$ is a quadratic <Galois character> unramified outside $S$. Its quadratic extension has a signed <square-free integer> representative $d=\pm\prod_{p\in S}p^{e_p}$, $e_p\in\{0,1\}$: an odd valuation outside $S$ would ramify there. There are at most $2^{s+1}$ choices, including the trivial character. Hence the pair of characters gives at most $2^{2s+2}$ cocycles. Each prime in $S$ is at most $2m$, because the three root differences have absolute values at most $2m$ and $2\leq2m$. Among the integers up to $2m$ there are at most $m$ primes: two, together with at most $m-1$ odd candidates. The <rational halving quotient bound from root size> is therefore
$$
\boxed{\#\bigl(E(\mathbb Q)/2E(\mathbb Q)\bigr)\leq2^{2s+2}\leq2^{2m+2}.}
$$
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