Let $X$ a sheaf and $ \{\mathcal{F}_{\lambda} \}_{\lambda}$ a arbitrary family of $ \mathcal{O}_X$-Modules. Because thats an abelian category it contains products and coproducts (=sums).
We considering the product presheaf $\prod \mathcal{F}_{\lambda}$ by defining $U \to \prod \mathcal{F}_{\lambda}(U)$ for euch open $U$ (resp. the sum presheaf $\bigoplus \mathcal{F}_{\lambda}$ by defining $U \to \bigoplus \mathcal{F}_{\lambda}(U)$).
My question is why the product presheaf as defined above is automatically a sheaf (therefore the sheaf axiom holds) where the sum is only a presheaf (therefore have to be sheafificated)?
Garbificated ? Did you try to prove that the product is a sheaf ?
Here is a counter-example for the direct sum : let $X=\mathbb{R}$ and $U=\mathbb{R}\setminus\mathbb{Z}$. Then $U$ is the disjoint union of the intervals $(n,n+1)$.
Let $\mathcal{F}_n=\mathbb{Z}$ be the constant sheaf and $s_n$ be the section of the presheaf $\bigoplus\mathcal{F}_k$ on $(n,n+1)$ such that every component of $s_n$ is zero except the $n$-th component which is $1$. In other words, $s_n\in(\bigoplus\mathcal{F}_k)((n,n+1))=\bigoplus\mathbb{Z}$ is the sequence $(...,0,0,1,0,0,...)$ where $1$ is at the $n$-th position.
I claim that these sections can't be glued to a section on $U$ (though they vacuously agreed on overlaps). If you want to glue these sections, you will need a section which have an infinite number of non zero component. And this won't be an element of $\bigoplus \mathcal{F}_n$.