Direct Sum/Product of Flat Modules Let $\left\{M_{\lambda}\right\}$ be a family of flat $A$-modules and define $M = \bigoplus_{\lambda} M_{\lambda}$. Let $0 \rightarrow N' \rightarrow N$ be an exact sequence of $A$-modules. Tensoring with $M$ gives $N' \otimes M \rightarrow N \otimes M$ or $\bigoplus_{\lambda} [N' \otimes M_{\lambda}] \rightarrow \bigoplus_{\lambda} [N \otimes M_{\lambda}]$. Each of the coordinate maps is a monomorphism, hence we actually have an exact sequence $0 \rightarrow \bigoplus_{\lambda} [N' \otimes M_{\lambda}] \rightarrow \bigoplus_{\lambda} [N \otimes M_{\lambda}]$ and so $\bigoplus_{\lambda} M_{\lambda}$ is flat. 
Question 1: Is the above argument accurate?
Question 2: Does the above argument work for the direct product $\prod_{\lambda} M_{\lambda}$ as well?
 A: Q1: Yes. I would write down the proof more concisely as follows: $M \otimes - \cong \oplus_i( M_i \otimes -)$ is a direct sum of exact functors, hence also exact (here we use what is called AB 4 for general abelian categories).
Q2: The tensor product does not commute with infinite products of modules. Therefore the proof breaks down here. And in fact, flat modules don't have to be closed under infinite products. The whole story can be found in section §4F of Lam's Lectures on Modules and Rings. There is the following result due to Chase:

Theorem. If $R$ is a ring, then the following are equivalent:
  
  
*
  
*$R$ is left coherent (i.e. every finitely generated left ideal is finitely presented).
  
*For every set $I$ the right $R$-module $R^I$ is flat.
  
*Flat right $R$-modules are closed under infinite products.
  

An inspection of the proof shows more precisely: Let $R$ be a ring which is not left coherent, say that $J$ is a left ideal which is not finitely presented. Choose an exact sequence $0 \to K \to F \to J \to 0$, where $F$ is a finitely generated free left $R$-module and $K$ is some left $R$-module which is not finitely generated. Let $I$ be the underlying set of $K$. Then the right $R$-module $R^I$ is not flat.
For example, the commutative ring $R := \mathbb{Q}[y,x_0,x_1,\dotsc]/(y x_i)_{i \geq 0}$ is not coherent, with $J=Ry$, $F=R$, $K=\sum_i R x_i$. Since $K$ is countable, it follows that $R^{\mathbb{N}}$ is not flat over $R$.
