Fix a prime p which doesn't divide the degree of K over ${\mathbb Q}$, and let ${\mathcal O}$ denote the ring of integers of ${\mathbb Q}_p(\chi)$ i.e. an extension of ${\mathbb Q}_p$ containing the values of $\chi$. Then the group algebra ${\mathcal O}[G]$ decomposes into a direct sum of 1-dimensional pieces over ${\mathcal O}$, one for each power of $\chi$.

Then $Sha(E/K)[p^\infty] \otimes {\mathcal O}$ being an ${\mathcal O}[G]$-module inherits such a decomposition. Concretely, the $\chi^i$-component of $Sha(E/K)[p^\infty] \otimes {\mathcal O}$ is the subset where $G$ acts by $\chi^i$.

This $\chi^i$-component is then a reasonable candidate to compare to the $p$-adic valuation of the algebraic part of $L(E,\chi^i,1)$.

Some further comments:

-Note that extending scalars to ${\mathcal O}$ increases the size of the modules so this has to be taken into account.

-The component corresponding to the trivial character is the invariants under $G$, and when $G$ has size prime-to-p this is simply $Sha(E/{\mathbb Q})[p^\infty] \otimes {\mathcal O}$ (which is good).

-To make a precise relationship between the $L$-value and Sha, you need to take into account the other terms in BSD. Namely:

*The torsion-term should work out exactly as above (decomposing into $\chi$-components).

*The periods have to be considered (which was ignored above in my vague phrase "the algebraic part of").

*The Tamagawa numbers give me pause -- possibly there is an analogous $\chi$-decomposition, but I don't see it now.

*Lastly, if K is ramified over ${\mathbb Q}$ then the discriminant of K appears in the BSD quotient (in the denominator which increases the size of Sha). To handle this, I imagine what should be done is that rather then considering the L-value alone, consider the L-value times the Gauss sum of the character. (By the conductor-discriminant formula this should give exactly the extra powers of p needed.)