Given an index set $I$ and $R$-modules $\{M_i\}_{i\in I}$, with $R$ being a commutative ring with identity, the direct product of $M_i$ is defined as the set $$\oplus_{i\in I}M_i =\{(m_i)_{i\in I}: m_i \in M_i \text{ such that all except finitely many $m_i = 0$}\}$$
Since for a tuple $(m_i)_{i\in I} \in \oplus_{i\in I}M_i$, not all elements are zero, what will be the zero elements in that group? since the direct sum of modules is a module.
What confuses me also is the fact that if $I$ is a finite set then $\oplus_{i\in I}M_i = \prod_{i\in I} M_i$, where $\prod_{i\in I} M_i$ is the direct product. For instance, Taking $\mathbb{Z}$-module $2\mathbb{Z}$ and $3\mathbb{Z}$, we have $2\mathbb{Z} \oplus 3\mathbb{Z} = \{(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)\}$
and in that case, we have an element, $(0,0)$, where all components are zero.
Best Answer
I think you are misunderstanding the phrase "all except finitely many are zero". It does not mean "not all of them are zero"; rather, it means "the nonzero elements are finite in number". That is to say, there may be 5 nonzero elements, or 100 of them, or 10 billion of them, but there can't be infinitely many of them.
Notice that this definition allows for the possibility that there may be no nonzero elements at all, because $0$ is a finite number.