The coordinate maps are also called projection maps. They send an element of a product set into the $\alpha$-th "coordinate" space. For instance, if $X = A\times B$, then $\pi_A:X\to A$ would be defined by $\pi(a,b) = a$ for all $(a,b)\in A\times B$. Countable products can be thought of similarly, since you can write elements as infinite sequences $(a_1, a_2, \ldots)$. Uncountable products are a little bit more unwieldy.
Let's also use this example to get some feel for the product $\sigma$-algebra. For a finite product $X = A\times B$, where now $A, B = \mathbb{R}$ and each is equipped with the Borel $\sigma$-algebra. Then the product $\sigma$-algebra on $X$ is generated by
$$\{\pi_A^{-1}(E): E\in \mathscr{B}_A\}\cup\{\pi_B^{-1}(F):F\in\mathscr{B}_B\}.$$
For any $E\in\mathscr{B}_A$, we have $$\pi_A^{-1}(E) = E\times\mathbb{R}.$$ Similarly $$\pi_B^{-1}(F) = \mathbb{R}\times F.$$
Since the product $\sigma$-algebra is closed under finite intersections, this means $E\times F \in\mathcal{M}_{A\times B}$. So the product $\sigma$-algebra contains all sets of the form $E\times F$, where $E\in\mathcal{M}_A$ and $F\in\mathcal{M}_B$. (These $\sigma$-algebras are the Borel $\sigma$-algebras in this example.) In particular, taking $A$ and $B$ to be intervals (open, half-open, closed, whatever) we see that $\mathcal{M}_{A\times B}$ contains all rectangles. Since the Borel $\sigma$-algebra on $\mathbb{R}^2$ can be generated by half-open rectangles, this shows you that the product $\sigma$-algebra contains the Borel $\sigma$-algebra. This is one of the things that we expect out of the Lebesgue $\sigma$-algebra on $\mathbb{R}^2$, once we get around to defining it.
This isn't a complete description of the product $\sigma$-algebra. In general, there are many more sets in the product $\sigma$-algebra than just the Cartesian products of measurable sets. But I hope this example gives you some insight into how the product $\sigma$-algebra works - it is not as intimidating as Folland makes (arguably everything) seem.
Please note that, even if $A$ is finite, an element of $\prod _{\alpha \in A}M_\alpha$ is not a subset of $\prod _{\alpha \in A}X_\alpha$. So it can not be (or generate) a $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$. I think you meant to write: the $\sigma$-algebra generated by $\{\prod _{\alpha \in A} E_\alpha : E_\alpha \in M_\alpha, \alpha \in A\}$.
Let $M$ be the $\sigma$-algebra generated by
$$
\{\pi_\alpha^{-1}(E_\alpha): E_\alpha \in M_\alpha, \alpha \in A\}
$$
and $N$ be the $\sigma$-algebra generated by
$$
\{\prod _{\alpha \in A} E_\alpha : E_\alpha \in M_\alpha, \alpha \in A\}
$$
First, as Tyr mentioned, if $A$ is (at most) countable, then $M=N$. In the general case, we have $M\subseteq N$. In all cases, the product $\sigma$-algebra is the SMALLEST $\sigma$-algebra that makes all the coordinate maps $\pi _\alpha$ measurables, and that is $M$. It is similar to the way we define product topology.
Why we want the product $\sigma$-algebra to be the SMALLEST $\sigma$-algebra that makes all the coordinate maps $\pi _\alpha$ measurables ?
Because the smallest $\sigma$-algebra that makes all the coordinate maps $\pi _\alpha$ measurables is precisely the only $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$ which makes the following proposition hold:
Let $(X,\Sigma)$ be a mensurable space and $f: X \to \prod _{\alpha \in A}X_\alpha$ then: $f$ is measurable IF AND ONLY IF , for all $\alpha \in A$, $\pi_\alpha \circ f$ is measurable.
Any $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$ larger than $M$ (the product $\sigma$-algebra) will result in the existence of $f$ which is not measurable but, for all $\alpha\in A$, $\pi_\alpha \circ f$ is measurable.
Any $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$ smaller than $M$ (the product $\sigma$-algebra) will result in the existence of a measurable $f$ such that there is $\alpha\in A$, $\pi_\alpha \circ f$ is not measurable.
In fact, let $S$ be a $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$.
If $S \varsupsetneq M$, take $X= \prod _{\alpha \in A}X_\alpha$ and $\Sigma=M$, then $Id: (\prod _{\alpha \in A}X_\alpha,M) \to (\prod _{\alpha \in A}X_\alpha, S) $ is NOT measurable, but, for all $\alpha\in A$, $\pi_\alpha \circ Id$ is measurable.
If $S \varsubsetneq M$, take $X= \prod _{\alpha \in A}X_\alpha$ and $\Sigma=S$, then $Id: (\prod _{\alpha \in A}X_\alpha,S) \to (\prod _{\alpha \in A}X_\alpha, S) $ is measurable, but there is $\alpha\in A$ such that $\pi_\alpha \circ Id$ is not measurable.
Best Answer
We can write
\begin{align} \bigl\{ \pi_\alpha^{-1}(E_\alpha) : E_\alpha \in \mathcal{M}_\alpha, \alpha \in \{1,2\}\bigr\} &= \{ \pi_1^{-1}(E_1) : E_1 \in \mathcal{M}_1\} \cup \{ \pi_2^{-1}(E_2) : E_2 \in \mathcal{M}_2\}\\ &= \{ E_1 \times X_2 : E_1\in \mathcal{M}_1\} \cup \{ X_1 \times E_2 : E_2 \in \mathcal{M}_2\}. \end{align}
In this form it is clear that this generating set is contained in
$$\{ E_1 \times E_2 : E_\alpha \in \mathcal{M}_\alpha\},$$
and on the other hand, every set in the latter generating family is the intersection of two members of the former, so the two families generate the same $\sigma$-algebra.