I'll address the question of generalising the formula $\mathrm{rank}(E/K) = \mathrm{rank}(E/\mathbb Q) + \mathrm{rank}(E_D/\mathbb Q)$.
In general, if $E/K$ is an elliptic curve, then there is an abelian variety
$$A := \mathrm{Res}_{K/\mathbb Q}(E)$$
called the restriction of scalars of $E$ with the property that
$$A(\mathbb Q) \cong E(K).$$
If $K/\mathbb Q$ is quadratic, then there is an isogeny $A\simeq E \times E_D$ and, since the rank is an isogeny invariant, you recover your formula.
If $K = \mathbb Q(2^{1/ 3})$, then $A$ is isogenous to $E\times B_2$, where $B_2$ is an abelian surface over $\mathbb Q$ that can be described as follows. Let $B = \mathbb Z[\zeta_3]\otimes_{\mathbb Z} E$, an abelian surface over $\mathbb Q$ with $\mathrm{Aut}_{\overline{\mathbb Q}}(B) \supset \langle \zeta_3\rangle$. In particular, $B$ admits cubic twists, and $B_2$ is the cubic twist corresponding to the extension $\mathbb Q(2^{1/ 3})$.
A lot of this and more can be deduced from this paper of Mazur–Rubin–Silverberg.
For a general number field, this is not always true: see [1], which (among studying the phenomenon in general), gives the example of $E:y^2 = x^3+\frac54 x^2 -2x-7$ over $\mathbb Q(\sqrt[6]{-11})$ as an example of a curve for which $E_D$ has negative root number for every D. Assuming BSD, this would mean that the rank of $E_D$ is never 0.
The more general question of the distribution of ranks of quadratic twists over arbitrary number fields has been studied by Klagsbrun–Mazur–Rubin [2]: see, for example, Conj. 7.12, which gives a precise conjecture for the average rank of $E_D$ (as $D$ varies over $K^\times/K^{\times 2}$), generalising Goldfeld's conjecture. I'm not sure if examples such as the one above exist when $K = \mathbb Q(\sqrt{l_1}, \ldots,\sqrt{l_n})$, but this paper would be a good place to start.
On the other hand, a positive answer to your question is well-known when $K=\mathbb Q$: the existence of a single $D$ was first proven in [3] on the analytic side (see [4] for a proof for infinitely many $D$), and the algebraic side follows from Wiles' modularity theorem + Gross–Zagier's proof of rank 0 BSD. More recently, the work of Alex Smith has proven Goldfeld's conjecture almost unconditionally.
For general number fields, if you insist on only considering $E_D$ where $D$ is an integer, than I'm not really sure what the answer should be: the set of $E_D$ is an extremely sparse subset of the set of quadratic twists, where $D$ should vary over $K^\times/K^{\times 2}$.
On the other hand, if you're happy to vary $D$ over $K^\times/K^{\times 2}$, then Smith's methods almost certainly extend to general number fields as well, at least to some extent. So I'd imagine that, in cases where your statement is true, a proof is probably within reach.
[1] T. Dokchitser, V. Dokchitser, Elliptic curves with all quadratic twists of positive rank. Acta
Arith. 137 (2009)
[2] Klagsbrun, Zev, Barry Mazur, and Karl Rubin. "Disparity in Selmer ranks of quadratic twists of elliptic curves." Annals of Mathematics (2013): 287-320.
[3] Bump, Daniel, Solomon Friedberg, and Jeffrey Hoffstein. "Nonvanishing theorems for L-functions of modular." Invent. math 102 (1990): 543-618.
[4] K. Ono and C. Skinner, ‘Non-vanishing of quadratic twists of modular 𝐿-functions’, Invent. Math. 134(3) (1998),
651–660.
Best Answer
You were overthinking this.
By Mordell-Weil we know that $E(K)$ is a finitely generated abelian group. Because $P^{\sigma}=P$ for all $P\in E(K)$, it follows that $[2]E(K)$ is contained in the image of the trace $T:E(L)\to E(K)$.
By virtue of $E(K)$ being finitely generated, already $E(K)/[2]E(K)$ is a finite elementary abelian $2$-group. Consequently, the same holds for its quotient group $E(K)/T(E(L))$. After all, we are moding out a subgroup containing $[2]E(K)$.