Is it true that $I\cap J\subset IJ\subset I+J\subset I\cup J$
If $R$ is a commutative ring and $I,J$ are any ideals of $R$, I don't know how the product is usually defined but I think for $IJ$ is the set of all finite sums with $\sum\limits_{k,n}i_kj_n$ with $i_k\in I, j_n\in J$, what is the sum of $2$ ideals, is it the same as the product with multiplication replaced by addition ?
No, $IJ\subseteq I\cap J\subseteq I\cup J\subseteq I+J$.
The middle inclusion is true for all sets.
So you just need to prove that $IJ\subseteq I\cap J$ and $I\cup J\subseteq I+J$.
It's easy to find counter-examples in $\mathbb Z$ to your inclusions:
$$(6\mathbb Z)(4\mathbb Z)=24\mathbb Z\subsetneq 12\mathbb Z=6\mathbb Z\cap 4\mathbb Z$$
$$6\mathbb Z\cup 4\mathbb Z \subsetneq 6\mathbb Z+4\mathbb Z=2\mathbb Z$$
And $I+J=\{i+j\mid i\in I,j\in J\}$ is the smallest ideal containing both $I$ and $J$. In $\mathbb Z$, this corresponds to the "greatest common divisors" of the generators of $I$ and $J$.
($I\cap J$ corresponds to the least common multiple.)