Galois group of the product of two fields being the product of Galois groups?

Yes, your theorem and its proof are correct. This is Theorem 1.14 in Chapter 6 of Lang's Algebra.

Your application to $\mathbb{Q}(\sqrt{a_1},\ldots,\sqrt{a_n})$ is also correct - by applying Theorem 1.14 repeatedly, one gets that $\text{Gal}(\mathbb{Q}(\sqrt{a_1},\ldots,\sqrt{a_n})/\mathbb{Q})$ injects into $\prod_{i=1}^n\text{Gal}(\mathbb{Q}(\sqrt{a_i})/\mathbb{Q})\cong C_2^n$.

The term you are looking for is compositum. If $K$ is a field and $E_1, E_2 \subset \bar{K}$ are two extensions of $K$, then their compositum $E_1 E_2$ is the subfield of $\bar{K}$ generated by $E_1$ and $E_2$. Note that unless $E_1, E_2$ are normal extensions the compositum is not uniquely specified by what $E_1, E_2$ look like as abstract extensions; you really do need to specify an embedding in some bigger field, which might as well be the algebraic closure for algebraic extensions.

The compositum of a Galois extension and another extension is Galois, so in particular if $E_1, E_2$ are both Galois then $E_1 E_2$ is Galois. Now there is an obvious embedding $\text{Gal}(E_1 E_2 / K) \to \text{Gal}(E_1 / K) \times \text{Gal}(E_2 / K)$ whose image is precisely the subgroup of elements which fix $E_1 \cap E_2$ (which is also Galois). When $E_1 \cap E_2 = K$ the conclusion follows.