Which groups does $U(2)$ cover? I can guarantee that $SU(2)$ covers only itself and $SO(3)$ because its center is the group $\{\pm I\}$. Then, as the kernel of a covering map of Lie groups is a discrete central subgroup, it must be a subgroup of $\{\pm I\}$.
But the center of $U(2)$ is $S^1\cdot I$. Thus, the discrete central subgroups of $U(2)$ are the cyclic groups defined by the roots of unity. But given such a cyclic group can one guarantee that there is a covering map from $U(2)$ whoose kernel is this cyclic group?
As the center of $U(2)$ is a circle, the base of this covering must have a central circle as well.
 A: Thinking in terms of $U(2)$ alone is confusing me a little so I am going to pull back to a double cover. $U(2)$ admits $U(1) \times SU(2)$ as its unique double cover, with covering map
$$U(1) \times SU(2) \ni (z, X) \mapsto zX \in U(2).$$
as you mentioned in the comments, where we implicitly use the embedding of $U(1)$ into $U(2)$ as the scalar multiples of the identity. The kernel of this map is the diagonal copy of $\mathbb{Z}/2$ generated by $(-1, -1)$. Now we can understand the quotient of $U(2)$ by the discrete central copy of $\mathbb{Z}/n$ by pulling it back to this double cover; if $n$ is odd the corresponding subgroup of $U(1) \times SU(2)$ is $\mathbb{Z}/(2n)$, embedded as
$$\mathbb{Z}/(2n) \ni k \mapsto (\exp \left( \frac{2 \pi i k}{n} \right), (-1)^k) \in U(1) \times SU(2)$$
and if $n$ is even the corresponding subgroup is $\mathbb{Z}/n \times \mathbb{Z}/2$, embedded as
$$\mathbb{Z}/n \times \mathbb{Z}/2 \ni (k, \pm 1) \mapsto (\exp \left( \frac{2 \pi i k}{n} \right), \pm 1) \in U(1) \times SU(2).$$
It follows that when $n$ is even the quotient is isomorphic to $U(1) \times SO(3)$ as you said in the comments. When $n$ is odd we can perform the quotient by first quotienting by the element $2 \in \mathbb{Z}/(2n)$, which allows us to ignore the $(-1)^k$ bit and gives another copy of $U(1) \times SU(2)$, then quotient by $1 \in \mathbb{Z}/(2n)$, which now acts by $(-1, -1)$, so we get a group isomorphic to $U(2)$.
I have to admit that this is a surprise to me! I was misled by an analogy to the group $SL_2(\mathbb{R})$ which turned out to be invalid. I really am actually surprised that $U(2)$ covers itself nontrivially.
