It's possible to generalize the counterexample from @user26857 of $(\mathfrak{a}:\mathfrak{b_1}) + (\mathfrak{a}:\mathfrak{b}_2) \subsetneq (\mathfrak{a} : \mathfrak{b}_1 \cap \mathfrak{b}_2)$ to get a counterexample where $\sum_{i \in I} (\mathfrak{a}: \mathfrak{b}_i) \subsetneq (\mathfrak{a}: \bigcap_i \mathfrak{b}_i)$ for an index set $I$ of any cardinality, including infinite.
Let $S$ be an arbitrary set with cardinality $|S|=|I|$ (i.e. possibly infinite), and let $k[S]$ denote the commutative ring with unit of polynomials with variables in $S$ and coefficients in the field (or more generally commutative ring with unit) $k$. I.e. finite linear combinations of finite formal products of elements of $S$.
Let $\mathfrak{a}$ denote the ideal consisting of "polynomials whose monomial summands are mixed-variable", in other words polynomials of degree $\ge 2$ that equal neither $\kappa S_i^n$ for some $\kappa \in K$, $S_i \in S$ (i.e. $i \in I$) nor any linear combination thereof (e.g. $\kappa_1 S_{i_1}^{n_1} + \kappa_2 S_{i_2}^{n_2}$). I believe that it is correct to write that
$$ \mathfrak{a} = \sum_{i \not= j,\\i,j \in I} \langle S_i S_j \rangle \,,$$
although please correct me if that is wrong. Now define $R = k[S] / \mathfrak{a}$.
Then for any $i \in I$, $\newcommand{\Ann}{\operatorname{Ann}}$ $\Ann(S_i) = (\langle 0 \rangle : S_i)$ in $R$ is (I think) $\sum_{j\not=i} \langle S_j \rangle$. This is since, given any $g \in \Ann(S_i)$, one must have that $gf \in \mathfrak{a}$ for all $f \in \langle S_i \rangle$, i.e. all $f = \kappa S_i$ for some $\kappa \in k$. Thus none of the monomial terms of $g$ can be any scalar multiple of a power of $S_i$ nor a constant, thus $g \in \sum_{j \not= i} \langle S_j \rangle$. So $\Ann (S_i) \subseteq \sum_{j \not= i} \langle S_j \rangle$, and the converse inclusion $\sum_{j \not= i} \langle S_j \rangle \subseteq \Ann(S_i)$ is even more immediate.
The point is that: $$ \sum_{i \in I} ( \langle 0 \rangle: \langle S_i \rangle) = \sum_{i \in I} \Ann(S_i) = \sum_{i \in I} \sum_{j \not= i, \\ j \in I} \langle S_j \rangle = \sum_{j \in I} \langle S_j \rangle \subsetneq R = \left( \langle 0 \rangle : \bigcap_{i \in I} \langle S_i \rangle \right) \,.$$
This is since:
Case I: $|S| = \infty$. Then $\bigcap_i \langle S_i \rangle = \langle 0 \rangle$, since by definition of $k[S]$ no element of $k[S]$ can have all $S_i$ be factors for all $i \in I$, and $\langle 0 \rangle$ in $k[S]$ is mapped to $\langle 0 \rangle$ in $k[S] / \mathfrak{a} = R$ via the order-preserving correspondence between the ideals of $k[S]$ and $k[S]/\mathfrak{a}$ (since obviously $\langle 0 \rangle \subseteq \mathfrak{a}$ in $k[S]$).
Case II: $|S| < \infty$. (OK, I think we also need that $|S| \ge 2$, as in user26857's counterexample, but whatever.) In this case we have $\bigcap_i \langle S_i \rangle = \langle S_1 S_2 \dots S_{i_{\max}} \rangle \subseteq \mathfrak{a}$ in $k[S]$ (at least when $|S| \ge 2$, although actually when $|S| < 2$, $\mathfrak{a}$ isn't even defined, right?, so whatever). Thus we again have that $\bigcap_i \langle S_i \rangle$ is $\langle 0 \rangle$ in $k[S]/\mathfrak{a}$, again due to the order-preserving correspondence between the ideals of $k[S]$ and $k[S]/\mathfrak{a}$.
In both cases, $\bigcap_i \langle S_i \rangle = \langle 0 \rangle$ in $k[S]/\mathfrak{a} = R$, and since $(\mathfrak{a}: \mathfrak{a}) = R$ for any ideal $\mathfrak{a}$, we have that $R =(\langle 0 \rangle : \langle 0 \rangle ) = (\langle 0 \rangle: \bigcap_i \langle S_i \rangle ) $, which allows us to conclude that in this case $\sum_{i \in I} \mathfrak{a} : \mathfrak{b}_i) \subsetneq (\mathfrak{a} : \bigcap_{i \in I} \mathfrak{b}_i )$.
Actually I don't think we even need to move to $k[S]/\mathfrak{a}$ from $k[S]$ for this counterexample to work. One has that $(\mathfrak{a} : \mathfrak{b} ) = R$ for any ideal $\mathfrak{b}$ such that $\mathfrak{b} \subseteq \mathfrak{a}$. This is since $(\mathfrak{a}: \mathfrak{b_2} ) \subseteq (\mathfrak{a} : \mathfrak{b_1})$ whenever $\mathfrak{b_1} \subseteq \mathfrak{b_2}$, and as mentioned above $(\mathfrak{a} : \mathfrak{a})$ (where here $\mathfrak{a}, \mathfrak{b}, \mathfrak{b}_1, \mathfrak{b}_2$ denote arbitrary ideals of an arbitrary commutative ring with unit $R$, sorry for abuse of notation). Therefore using this fact we can say, since in both cases above $\bigcap_{i \in I} \langle S_i \rangle \subseteq \mathfrak{a}$ in $k[S]$, that:
$$ \sum_{i \in I} (\mathfrak{a}: \langle S_i \rangle) = \sum_{i\in I } \sum_{j\not=i,\\j \in I} \langle S_j \rangle = \sum_{j \in I} \langle S_j \rangle \subsetneq k[S] = (\mathfrak{a} : \bigcap_{i \in I} \langle S_i \rangle) \,. $$
One can also get simple counterexamples to show that possibly $\sum_{i \in I} (\mathfrak{a}_i : \mathfrak{b}) \subsetneq (\sum_{i \in I} \mathfrak{a}_i : \mathfrak{b})$ using the same ring $R = k[S]$ and argument that $(\mathfrak{b}:\mathfrak{b}) = R$ for any ideal $\mathfrak{b} \subseteq R$. Again, regardless (I believe) of the cardinality of $S$ (OK, again, as long as it is $\ge 2$ at least). (Although I think one can still do something similar using $\langle 2 \rangle$ and $\langle X \rangle$ in $\mathbb{Z}[X]$ but again whatever.)
Basically the essential claim is that, for any $i \in I$,
$$\left(\langle S_i\rangle : \sum_{j \in I} \langle S_j \rangle \right) = \langle S_i \rangle \,.$$
Since $\mathfrak{a} \subseteq (\mathfrak{a} : \mathfrak{b})$ always for arbitrary ideals $\mathfrak{a}, \mathfrak{b}$, one direction of inclusion is obvious. Now assume that $f \in \left(\langle S_i\rangle : \sum_{j \in I} \langle S_j \rangle \right)$: then for all $g \in \sum_{j \in I} \langle S_j \rangle$, $fg = \tilde{f}_g S_i$ (i.e. $\in \langle S_i \rangle$) for some $\tilde{f}_g$ depending on $g$. However, whenever for $j \not=i$, $g = S_j \in \sum_{j \in I} \langle S_j \rangle$, this is only possible if $S_i$ is a factor of $f$, i.e. only if $f \in \langle S_i \rangle$. Since the condition has to hold for all $g \in \sum_{j \in I} \langle S_j \rangle$, it follows that our arbitrary $ f \in \left(\langle S_i\rangle : \sum_{j \in I} \langle S_j \rangle \right)$ is always such that $f \in \langle S_i \rangle$, proving the other inclusion and thus the purported equality above.
Using this claim, we get that (for any $k[S]$ with $|S| \ge 2$):
$$ \sum_{i \in I} \left( \langle S_i \rangle : \sum_{j \in I} \langle S_j \rangle \right) = \sum_{i \in I} \langle S_i \rangle \subsetneq R = k[S] = \left( \sum_{i \in I} \langle S_i \rangle : \sum_{j \in I} \langle S_j \rangle \right) \,,$$
since obviously $\sum_{i \in I} \langle S_i \rangle$ and $\sum_{j \in I} \langle S_j \rangle$ are the same ideal denoted differently and $(\mathfrak{b}: \mathfrak{b} ) = R$ for arbitrary ideals $\mathfrak{b} \subseteq R$ for arbitrary commutative rings with unit $R$. So again @user26857's suggestion to use examples where the right side equals $R$ is applicable here as well. Anyway the above is a counterexample where indeed $\sum_{i \in I} ( \mathfrak{a}_i : \mathfrak{b}) \subsetneq ( \sum_{i \in I} \mathfrak{a}_i : \mathfrak{b})$, and moreover it does not depend upon the cardinality of the index set $I$ being either finite or infinite.
I realized relatively recently that this question is actually related to a previous question of mine, since for the case that $|I|=2$, the properties I am referring to can be written $(\mathfrak{a}: \mathfrak{b}_1) + (\mathfrak{a}: \mathfrak{b}_2) \subseteq (\mathfrak{a} : \mathfrak{b}_1 \cap \mathfrak{b}_2)$ and $(\mathfrak{a}_1 : \mathfrak{b}) + (\mathfrak{a}_2 + \mathfrak{b}) \subseteq (\mathfrak{a}_1 + \mathfrak{a}_2 : \mathfrak{b})$, and these properties holding with equality (for finitely generated ideals, not arbitrary ideals) are equivalent conditions for being an arithmetical ring, see here or here for references which I found helpful.
