We need precisely two non-identity elements in $C_2\times C_2$ to generate the group. Fixing two non-identity generators for $V_4$, we need only figure out how many ways we can map those two to two distinct non-identity elements of $C_2$ (since a homomorphism is determined by what is done to generators, and we'll need to map the fixed generators of $V_4$ to two distinct non-identity elements of $C_2\times C_2$ to get an isomorphism).
Indeed, the answer is $6$. There are $3$ options for where we send the first generator, and regardless of which of those $3$ we choose, there $2$ remaining options for where we send the second. Thus, there are $3\cdot 2=6$ ways to make such a map.
See also this related answer.
Welcome to MSE!
Yes. There might be multiple bijections. As a simple example, look at
$$\mathbb{Z} / 2 \times \mathbb{Z} / 2 \cong \mathbb{Z} / 2 \times \mathbb{Z} / 2$$
We can use the identity, which is an isomorphism. We can also use the flippy isomorphism $(a,b) \mapsto (b,a)$ (which is its own inverse).
As for your second question, I'm not sure what you mean by whether the "kind of isomorphism picked does not affect how the structural similarity between $A$ and $B$ [is] preserved"...
I will say that no matter what isomorphism you use, all the group-theoretic properties will be the same. For instance, their orders, how many subgroups, whether there exist torsion elements, etc. So I'll say that as a good first answer: no, it doesn't matter at all which isomorphism you use.
Of course, like everything in math, this is only mostly true. In fact, different isomorphisms carry different information, and just because two groups are isomorphic doesn't always mean we want to consider them the same! A good example of this is the study of lattices in $\mathbb{R}^n$. These are all (abstractly) isomorphic to $\mathbb{Z}^n$, but it turns out to be a bad idea to identify them all as "the same".
There are also times when different isomorphisms carry different information. A famous example is the isomorphism of a (finite dimensional) vector space to its double dual. There are lots of ways to do this (abstractly) but it turns out there's one unique best choice of isomorphism. To formulate this properly we need the language of category theory, and I'm not sure if this is the right place to go into that. Rest assured, though, you'll hear more about it as you continue progressing in mathematics!
The long and the short of it is that every isomorphism will preserve all the group theoretic properties of your object of interest. However different choices of isomorphism can sometimes lead to counter intuitive statements (galois theory is rather infamous for this...). If there is a canonical way to make the isomorphism, rather than finding some random isomorphism by making arbitrary choices along the way (choosing a basis, etc.) that canonical isomorphism tends to be "better behaved" in a way that you'll learn about when you start learning about categories.
I hope this helps ^_^
Best Answer
We have here several isomorphisms. The first one, arising from CRT, is $$ U(20)\cong U(4) \times U(5), $$ because $\gcd(4,5)=1$. Then we have isomorphisms $$ U(4)\cong C_2,\quad U(5)\cong C_4. $$ Finally we have an isomorphism $$ U(16)\cong C_2 \times C_4. $$ Note that the isomorphisms $U(20)\cong C_2\times C_4$ and $U(16)\cong C_2\times C_4$ need not coincide.