So apparently there is some kind of a connection between quadratic characters and twists of elliptic curves that I can't understand.
I am well aware of quadratic twisting of an elliptic curve $E$, we get a curve $E^d$, and the curves are isomorphic over $\mathbb{Q}(\sqrt{d})$.
I also know about quadratic characters associated to fields $\mathbb{Q}(\sqrt{d})$, given by $$\chi=\chi_{\mathbb{Q}(\sqrt{d})}:\mathbb{Z}_{>0} \rightarrow \mathbb{C},$$ $$ \chi(n)=\left( \frac{disc(\mathbb{Q}(\sqrt{d}))}{n} \right). $$
In literature I often see twists of elliptic curves $E^d$ being denoted by $E^{\chi}$. Why is that so? what is the connection? Am I even using the right character here?
There are a few ways to intemperate this question. As @Mathmo123 says in their comment there is a natural bijection between quadratic characters and twists of $E$ (when the $j$-invariant is not $0$ or $1728$).
From a more elementary point of view we can define a quadratic twist of $E : y^2 = f(x)$ by $E^d : dy^2 = f(x)$. Notice that this is determined by $d$ up to squares - hence is determined by the quadratic extension $F = \mathbb{Q}(\sqrt{d}) / \mathbb{Q}$. We obtain a quadratic character $$ \chi : G_{\mathbb{Q}} \to \langle \pm 1 \rangle$$ of the absolute galois group $G_{\mathbb{Q}}$ by restricting $\sigma \in G_{\mathbb{Q}}$ to $F$ (note that $\operatorname{Gal}(F/\mathbb{Q}) \cong C_2$) (conversely we obtain a quadratic extension of $\mathbb{Q}$ as the fixed field of the kernel of such a character). Thus there is a bijective correspondence between quadratic twist of $E$ and quadratic characters of the absolute galois group.
A useful application of this thinking is the following
Lemma We have $$E^d(\mathbb{Q}) = \{P \in E : \sigma(P) = \chi(\sigma)P \}$$
Proof: Notice that we have $P \in E^d(\mathbb{Q})$ if and only if $P = (x', \sqrt{d}y')$, where $P' = (x', y') \in E(\mathbb{Q})$. Thus $\sigma(P) = P$ if $\sigma$ fixes $F$, or $\sigma(P) = -P$ otherwise, which is precisely the statement.