I want to show that the following set (given group structure with matrix multiplication)
$$D=\left\{ D(\theta)=\begin{pmatrix} e^{i\theta/2} & 0 \\ 0 & e^{-i\theta/2} \end{pmatrix}, 0 \leq \theta \leq 4\pi \right\}$$
covers twice $R$, the group of rotations in $\mathbb{R}^2$.
$$R=\left\{ R(\theta)=\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}, 0 \leq \theta \leq 4\pi \right\}$$
1) When we say that a group covers $n$ times another group means that there is an $1$ to $n$ homomorphism?
Let $f: D \to R$ so that $D(\theta)\to R(\theta)$ Because $0 \leq \theta \leq 4 \pi$ it happens that $D(\theta+2\pi)\to R(\theta)$.
2) Is $f$ the group homomorphism in order to prove that $D$ covers $R$?
3) When can be said that a group $G$ is the universal cover of a group $G'$?
One thing to note is that there seems to be some topological content to your question; it's not strictly group-theoretical. In particular the language "$A$ covers $B$ twice" (or, "$A$ is a double cover for $B$"), and especially "universal cover," are referencing covering space theory. Thus, the groups you're working with aren't just groups, but topological groups i.e. groups that are also topological spaces such that the group composition and inversion maps are continuous.
1) Yes, except I would say "$n$ to $1$", and I'd like to make the topological assumptions explicit. If we have topological groups $A,B$ and a homomorphism $A\rightarrow B$, this homomorphism is an "$n$-fold cover" if: a) it is surjective b) it is continuous as a topological map and c) each element of $B$ has a discrete set of $n$ preimages in $A$.
Because these are groups, we can check (c) just by looking at the size of the kernel.
If we weren't dealing with groups, we'd have more to check to say that we had an $n$-fold cover. In particular, we'd have to check that for each point of $B$, each of the $n$ preimages has a neighborhood such that the restriction of the map to that neighborhood is a homeomorphism. But since we are dealing with groups, this is taken care of as long as we know that the kernel is discrete.
2) Yes, that's the right homomorphism.
3) For any topological spaces (not just groups), $G$ is a "universal cover" of $G'$ if there is a covering map $G\rightarrow G'$ and $G$ is simply connected.