This is an "abstract nonsense" answer. If you're not familiar with a bit of category theory this will probably be useless.
For the second question :
Denote by $L$ the sheafification functor $\mathbf{Psh}(X) \to \mathbf{Sh}(X)$.
By definition (or construction,...) $L$ is left adjoint to the inclusion $i: \mathbf{Sh}(X) \to \mathbf{Psh}(X)$; hence it preserves colimits.
Now let $(F^l)_l$ be a family of presheaves, $\displaystyle\bigoplus_lF^l$ can be defined as in Malkoun's answer and is actually the coproduct of the $F^l$'s in $\mathbf{Psh}(X)$; hence $L(\displaystyle\bigoplus_lF^l) = \displaystyle\bigoplus_lL(F^l)$ which is exactly what you wanted.
Edit : there's a mistake in what comes right above : $L$ does preserve colimits, but the inclusion functor need not do so as well; hence the second $\bigoplus$ is not the direct sum in $\mathbf{Psh}(X)$, it is the direct sum in $\mathbf{Sh}(X)$; which can be defined using $L$. Hence the answer to your question is "yes if the second $\bigoplus$ is taken in $\mathbf{Sh}(X)$ and not in $\mathbf{Psh}(X)$.". In particular, taking sections over $U$ need not be the sum of the sections.
By the way you can answer the first question in the same way: $\mathbf{Ab}$ is an abelian cocomplete category, hence $\mathbf{Psh}(X)$ is also abelian cocomplete; and also for any open set $U\subset X$, the "evaluation at $U$" functor preserves colimits, hence $(\displaystyle\bigoplus_lF^l)(U)$ is $\displaystyle\bigoplus_lF^l(U)$. Restriction morphisms come from the universal property of the coproduct, as described in Makoun's answer ($\displaystyle\bigoplus_l$ can in fact be seen as a functor $\mathbf{Ab}^I \to \mathbf{Ab}$ where $I$ is the set of indices for this very reason)