Why is the arbitrary sum, but not the arbitrary intersection, of ideals an ideal?

I finally found a source (see here) which gives the definition: $$\sum_{j \in J} I_j := \left\{ \sum_{j\in J} x_j: x_j \in I_j\ (\forall j)\ \text{and only finitely many }x_j\text{ are nonzero} \right\} .$$

Related to Arnaud D.'s comment, we apparently have the relationship that: $$\sum_{j\in J} I_j = \left\langle \bigcup_{j\in J} I_j \right\rangle $$ where $\langle \rangle$ denotes the ideal generated by a set, the intersection of all ideals containing the set.

Thus the set of all proper ideals $I \subseteq R$ of a ring turns out to not only be a partial order under set inclusion, but actually a lattice, meaning that all supremums (least upper bounds) and infimums (greatest lower bounds) exist/are defined.

The infimum of a set of ideals is the arbitrary intersection of the ideals, while the supremum of ideals is the sum of ideals, since it is the smallest ideal containing all of the ideals. Thus the sum and intersection of ideals are indeed somewhat related operations.

Claim: $$\sum_{j\in J} I_j = \left\langle \bigcup_{j\in J} I_j \right\rangle $$ Proof: Let $ x \in \sum_{j\in J} I_j$. Then $x=x_1 + \dots + x_n$, with $x_1 \in I_{j_1}, \dots, x_n \in I_{j_n}$, $j_1, \dots, j_n \in J$. Then since $x_1, \dots, x_n \in \bigcup_{i=1}^n I_{j_i}$, we have that $x_1 + \dots + x_n \in L$ for any ideal $L$ such that $L \supseteq \bigcup_{i=1}^n I_{j_i}$. But of course because $\bigcup_{i=1}^n I_{j_i} \subseteq \bigcup_{j \in J} I_j$, so: $$\bigcup_{i=1}^n I_{j_i} \subseteq \bigcup_{j \in J} I_j \subseteq \bigcap_{J \supseteq \bigcup_{j \in J} I_j} J = \left\langle \bigcup_{j \in J} I_j\right\rangle, \\ L=\left\langle \bigcup_{j \in J} I_j\right\rangle \supseteq \bigcup_{i=1}^n I_{j_i}$$ so $x_1 + \dots + x_n \in \left\langle \bigcup_{j \in J} I_j\right\rangle$.

$x_1 + \dots + x_n \in \sum_{j\in J} I_j$ was arbitrary, we have shown that $\sum_{j\in J} I_j \subseteq \left\langle \bigcup_{j \in J} I_j\right\rangle$.

Since $\sum_{j \in J} I_j$ is an ideal and $\sum_{j \in J} I_j \supseteq \bigcup_{j\in J} I_j$, $$\sum_{j \in J} I_j \supseteq \bigcap_{\{J: J \supseteq \bigcup_{j\in J} I_j\}} J = \left(\bigcap_{\{J: J \supseteq \bigcup_{j \in J} I _j \} \setminus \{\sum_{j \in J} I_j \}} J\right) \cap \left( \sum_{j \in J} I_j \right).$$ However, $\bigcap_{\{J: J \supseteq \bigcup_{j\in J} I_j\}} J$ is literally the definition of $\left\langle \bigcup_{j\in J} I_j \right\rangle$, i.e. $\bigcap_{\{J: J \supseteq \bigcup_{j\in J} I_j\}} J = \left\langle \bigcup_{j \in J} I_j \right\rangle$ trivially, we have shown that $$\sum_{j \in J} I_j \supseteq \left\langle \bigcup_{j \in J} I_j \right\rangle$$ In other words, $\left\langle \bigcup_{j \in J} I_j \right\rangle$ is the smallest ideal containing $\bigcup_{j \in J} I_j$, $\sum_{j\in J}I_j$ is an ideal containing the union, so it must contain the smallest ideal containing the union. $\square$

Let $(I_j)_{j\in J}$ be a family of ideals of a commutative ring $R$. Their sum is the set of all finite sums $\sum_{k=1}^n a_k$ such that $a_k\in \bigcup_{j\in j}I_j$ for all $1\leq k\leq n$. This definition is equivalent to the second one you give, and you can check that it is indeed an ideal and that it is the smallest ideal containing all the $I_j$.

For your second question, the intersection $\bigcap_{j\in J} I_j$ is an ideal. Indeed, it is non-empty as $0\in I_j$ for all $j$; and if $a,b\in \bigcap_{j\in J} I_j$ and $r\in R$, then $a,b\in I_j$ for all $j$, so that $a+b\in I_j$ and $ra\in I_j$ for all $j$, and thus $a+b\in \bigcap_{j\in J} I_j$ and $ra\in \bigcap_{j\in J} I_j$.