I've found a proof of the structure of maximal abelian normal subgroups of an extraspecial group of order $2^{2m+1}$ in the book "Endlichen Gruppen I" by B. Huppert but there is a part of the proof I don't quite understand.
Assume $G=\mathcal{D_{1}}*\cdots*\mathcal{D_{m-1}}*\mathcal{Q}$ (the central product of $m-1$ dihedral groups of order $8$ and quaternion). Then takes $U$ a maximal abelian normal subgroup and decomposes the element $a\in U$ into two a product of two elements, i.e., $a=a_{1}a_{2}$ with $a_{1}\in\mathcal{D_{1}}*\cdots*\mathcal{D_{m-1}}$ and $a_{2}\in\mathcal{Q}.$ These elemets are uniquely determined up two a factor of $Z(G).$ Then the subgroups $U_{i}$ are defined as the "product of all $a_{i}", i=1,2$ respectively, where each of the $a_{i}$ picked in their respective groups, and this is the part I don't catch. Does it mean $U_{i}$ is generated by $a_{i}$? Is $U_{i}$ generated by the product of some $a_{i}^{(1)},a_{i}^{(2)},\ldots,a_{i}^{(r)}$? I don't understand how are defined the groups $U_{i}.$
I would greatly appreciate any help or idea.
For the first question, this is a classical result, You know that for any $a \in U$, $\langle a \rangle \leqslant G$, so by Lagrange, you get $$|\langle a \rangle | = o(a) \mid 2^{2m+1} |G|$$
Hence, $o(a)$ is a power of $2$. Then if $\forall a \in U, a^2 = e$ you have $a = a^{-1}$,thus you get for $a,b \in U, \, aba^{-1}b^{-1}= abab = (ab)^2 = e$, hence $U$ is abelian which is absurd.
You thus have some element $a \in U$ such that $a^2 \neq e$ hence $o(a) >2$, so since $o(a) = 2^k$ for some integer $k$, you have $k > 1$. You then need to consider $b= a^{2^{k-2}}$, for $l \in \mathbb{N}^*$, you have $b^l = a^{2^{k-2}l}$, hence if $l < 4$, $2^{k-2}l < 2^k$ and $b^l \neq e$.
And of course $b^4 = a^{2^k} = e$ so $o(b)=4$ and $b \in U$.
You could try to show this result: if $\forall x \in G, \, x^2 = e$, $G$ can be endowed with a $\mathbb{F}_2$-vector space structure. However you should neither not post too many question at a time, nor put links of articles, write them down.
For the conclusion of the last paragraph, $U_1 \times U_2$ is the pushout of the inclusions $Z(G) \to U_i$, this gives you the cardinality result