I found a case in which the (external) direct sum of R-modules do not satisfy the universal property of the direct product of R-modules.
However, I can't think of one in which the direct product of R-modules do not satisfy the universal property of the direct sum of R-modules....
Could anyone suggest any counter example?
Explicitly: consider the countable product of $\mathbb{Z}$, $\Pi_{i \in \mathbb{N}} \mathbb{Z}$, as a $\mathbb{Z}-$module. Let $\phi_i:\mathbb{Z} \rightarrow \mathbb{Z}$ be the identity map. If the product were to satisfy the universal property of direct sum, there would be a unique map $\phi:\Pi_{i \in \mathbb{N}}\mathbb{Z} \rightarrow \mathbb{Z}$ which is the identity on the coordinates. But it is not true that there is a unique map. For instance, there is no restriction on the value of $\phi(1,1,\ldots,1,\ldots)$
Any object satisfying a universal property is unique up to isomorphism.
A countable direct sum of $\mathbb{Z}$ is not isomorphic to a countable direct product of $\mathbb{Z}$ since they have different cardinalities. Hence neither satisfies the universal property of the other.
To expand on Seth's answer. In the case where the index set $I$ has finite cardinality $\bigoplus_{i\in I} M_i$ is the same as $\prod_{i\in I} M_i$. The cases where they differ are when you are forming a direct sum and direct product over an infinite index set.
In such cases $\bigoplus_{i\in I} M_i=\{(m_i)_{i\in I}\vert m_i\in M_i, m_i=0_{M_i}$ for all but finitely many $i\in I\}$. Another way of saying this is that an element in the direct sum is almost trivial.
On the other hand $\prod_{i\in I} M_i=\{(m_i)_{i\in I}\vert m_i\in M_i\}$. Notice the direct product lacks that restriction on its elements the direct sum has. Also, you should be able to see from these expressions why the direct sum and the direct product will coincide in the finite index case.
