Let $E_d$ be the group of rational points on the elliptic curve $$y^2=(x-d)x(x+d)$$
It is well-known – and easy to verify by elementary algebra – that $(x,y)\in E_d$ belongs to $2E_d$ precisely when all three of $x-d$, $x$, $x+d$ are squares.
This seems too neat to be a mere coincidence.
Is there a natural generalisation of this fact? Or a more conceptual way to understand why it ought to be true?
It looks as if you're working over $\mathbb Q$. And you are using an elliptic curve in which all of the 2-torsion is rational, so $E[2]$ is isomorphic to (say) $\boldsymbol\mu_2^2$ as a Galois module, where $\boldsymbol\mu_2=\{\pm1\}$ is the group of square roots of 1. Okay, now consider the injection (this comes from basic Kummer theory of elliptic curves) \begin{align*} E(\mathbb Q)/2E(\mathbb Q) &\hookrightarrow H^1(G_{\overline{\mathbb Q}/\mathbb Q},E[2]) \\ &\cong H^1(G_{\overline{\mathbb Q}/\mathbb Q},\boldsymbol\mu_2^2) \\ &\cong H^1(G_{\overline{\mathbb Q}/\mathbb Q},\boldsymbol\mu_2)^2 \\ &\cong \mathbb Q^*/(\mathbb Q^*)^2\times \mathbb Q^*/(\mathbb Q^*)^2. \end{align*} (The final isomorphism is standard Kummer theory.) So this at least tells you that whether $P$ is in $2E(\mathbb Q)$ should depend on whether certain values are squares. If you trace through all of the maps, which requires choosing a basis for $E[2]$, you'll find (for one of the choices) that it is given by $$ P \longmapsto \bigl(x(P),x(P)-d\bigr), $$ so $P$ is in $2E(\mathbb Q)$ if and only if both $x(P)$ and $x(P)-d$ are squares. Of course, from the equation of $E$, any two of $x(P)$, $x(P)-d$ and $x(P)+d$ being a square forces the third one to also be square. Finally, if $x(P)=0$ or $x(P)=d$, one finds during the analysis that one needs to use a different formula. For example, since $x$ and $(x-d)(x+d)$ differ multiplicatively by a square, one has $$ 0 \longmapsto (-d^2,-d) = (-1,-d) \in \mathbb Q^*/(\mathbb Q^*)^2\times \mathbb Q^*/(\mathbb Q^*)^2. $$
To answer your second question, if $E[m]\subset E(K)$, then one similarly gets an injection $$ E(K)/mE(K) \hookrightarrow K^*/(K^*)^m\times K^*/(K^*)^m, $$ so (more or less) there are rational functions $f$ and $g$ in $K(E)$ such that for $P\in E(K)$ we have $$ P \in mE(K) \;\Longleftrightarrow\; \text{$f(P)$ and $g(P)$ are $m$'th powers in $K^*$.} $$
A geometric view to complement Joe Silverman's cohomological answer:
Suppose more generally that $E$ is the "product-is-square" elliptic curve $$ y^2 = (x-e_1)(x-e_2)(x-e_3) $$ for some pairwise distinct $e_i$, and let $C$ be the "each-is-square" curve $$ \begin{cases} y_1^2 = x-e_1, \cr y_2^2 = x-e_2, \cr y_3^2 = x-e_3, \end{cases} $$ Then $E$ is a double cover of the $x$-line, and $C$ is a $\{\pm1\}^3$ cover with $E$ as an intermediate cover with Galois group isomorphic with $({\bf Z} / 2{\bf Z})^2$; explicitly the $4:1$ map $C \to E$ is $$(x,y_1,y_2,y_3) \mapsto (x, y_1 y_2 y_3).$$ By Riemann-Hurwitz, $E$, like $C$, has genus $1$, so this $({\bf Z} / 2{\bf Z})^2$ cover $C \to E$ is unramified. Hence if we choose for the origin of $C$ one of the preimages of the origin of $E$, the cover must be the doubling map.
