Idea of Covering Group 
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*$SU(2)$ is the covering group of $SO(3)$. What does it mean and does it have a physical consequence? 

*I heard that this fact is related to the description of bosons and fermions. But how does it follow from the fact that $SU(2)$ is the double cover of $SO(3)$?  
 A: Great, important question.  Here's the basic logic:


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*We start with Wigner's Theorem which tells us that a symmetry transformation on a quantum system can be written, up to phase, as either a unitary or anti-unitary operator on the Hilbert space $\mathcal H$ of the system.

*It follows that if we want to represent a Lie group $G$ of symmetries of a system via transformations on the Hilbert space, then we must do so with a projective unitary representation of the Lie group $G$.  The projective part comes from the fact that the transformations are unitary or anti-unitary "up to phase,"  namely we represent such symmetries with a mapping $U:G\to \mathscr U(\mathcal H)$ such that for each $g_1,g_2\in G $, there exists a phase $c(g_1, g_2)$ such that
\begin{align}
  U(g_1g_2) = c(g_1, g_2) U(g_1) U(g_2)
\end{align}
where $\mathscr U(\mathcal H)$ is the group of unitary operators on $\mathcal H$.  In other words, a projective unitary representation is just an ordinary unitary representation with an extra phase factor that prevents it from being an honest homomorphism.

*Working with projective representations isn't as easy as working with ordinary representations since they have the pesky phase factor $c$, so we try to look for ways of avoiding them.  In some cases, this can be achieved by noting that the projective representations of a group $G$ are equivalent to the ordinary representations of $G'$ its universal covering group, and in this case, we therefore elect to examine the representations of the universal cover instead.

*In the case of $\mathrm{SO}(3)$, the group of rotations, we notice that its universal cover, which is often called $\mathrm{Spin}(3)$, is isomorphic to $\mathrm{SU}(2)$, and that the projective representations of $\mathrm{SO}(3)$ match the ordinary representations of $\mathrm{SU}(2)$, so we elect to examine the ordinary representations of $\mathrm{SU}(2)$ since it's more convenient.  
This is all very physically important.  If we had only considered the ordinary representations of $\mathrm{SO}(3)$, then we would have missed the "half-integer spin" representations, namely those that arise when considering rotations on fermionic systems.  So, we must be careful to consider projective representations, and this naturally leads to looking for the universal cover.
Note: The same sort of thing happens with the Lorentz group in relativistic quantum theories.  We consider projective representations of $\mathrm{SO}(3,1)$ because Wigner says we ought to, and this naturally leads us to consider its universal cover $\mathrm{SL}(2,\mathbb C)$.
A: After the answers by joshphysics and user37496, it seems to me that a last remark remains. 
The quantum relevance of the universal covering Lie group in my opinion is (also) due to a fundamental theorem by Nelson. That theorem  relates Lie algebras of symmetric operators with unitary representations of a certain Lie group generated by those operators. The involved Lie group, in this discussion, is always a universal covering.
In quantum theories one often encounters a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ such that:
(1) They  are symmetric (i.e. defined on a dense domain $D(A_i)\subset {\cal H}$ where $\langle A\psi|\phi\rangle = \langle \psi|A\phi\rangle$)
and
(2) they enjoy the commutation relations  of some Lie algebra  $\ell$:
$$[A_i,A_j]= \sum_{k=1}^N iC^k_{ij}A_k$$
on a common invariant domain ${\cal D}\subset {\cal H}$.
As is known, given an abstract Lie algebra $\ell$ there is (up to Lie group isomorphisms) a unique simply connected Lie group ${\cal G}_\ell$ such that its Lie algebra coincide with $\ell$. ${\cal G}_\ell$ turns out to be the universal covering of all the other Lie groups whose Lie algebra is $\ell$ itself.
All those groups, in a neighbourhood of the identity are isomorphic to a corresponding neighbourhood of the identity of ${\cal G}_\ell$. (As an example just consider the simply connected $SU(2)$ that is the universal covering of $SO(3)$) so that they share the same Lie algebra and are locally identical and differences arise far from the neutral element. 
If (1) and (2) hold, the natural question is:
Is there a strongly continuous unitary representation ${\cal G} \ni g \mapsto U_g$ of some Lie group  $\cal G$ just admitting $\ell$ as its Lie algebra, such that 
$$U_{g_i(t)} = e^{-it \overline{A_i}}\:\: ?\qquad (3)$$
Where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of $\cal G$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is some self-adjoint extension of $A_i$.
If it is the case, $\cal G$ is a continuous symmetry group for the considered physical system, the self adjoint operators $\overline{A_i}$ represent physically relevant observables. If time evolution is included in the center of the group (i.e. the Hamiltonian is a linear combination of the $A_i$s and commutes with each of them) all these observables are conserved quantities.
Otherwise the situation is a bit more complicated, nevertheless  one can define conserved quantities parametrically depending on time and belonging to the Lie algebra of the representation (think of the boost generators when $\cal G$ is $SL(2,\mathbb C)$). 
Well, the fundamental theorem by Nelson has the following statement.
THEOREM (Nelson)
Consider a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ satisfying (1) and (2) above. If ${\cal D}$ in (2) is a dense subspace such that the symmetric operator
$$\Delta := \sum_{i=1}^N A_i^2$$
is essentially self-adjoint on $\cal D$ (i.e. its adjoint is self-adjoint or, equivalently, $\Delta$ admits a unique self-adjoint extension, or equivalently its closure $\overline{\Delta}$ is self-adjoint), then:
(a) Every $A_i$ is essentially self-adjoint on $\cal D$,
and
(b) there exists a strongly continuous unitary representation on $\cal H$ of the unique simply connected Lie group ${\cal G}_\ell$ admitting $\ell$ as Lie algebra, completely defined by the requirements:
$$U_{g_i(t)} = e^{-it \overline{A_i}}\:\:,$$
where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of ${\cal G}_\ell$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is the unique  self-adjoint extension of $A_i$ coinciding to $A_i^*$ and with the closure of $A_i$.
Notice that the representation is automatically unitary and not projective unitary: No annoying phases appear.
The simplest example is that of operators $J_x,J_y,J_z$. It is easy to prove  that $J^2$ is essentially self adjoint on the set spanned by vectors $|j,m, n\rangle$. The point is that one gets this way unitary representations of $SU(2)$ and not $SO(3)$, since the former is the unique simply connected Lie group admitting the algebra of $J_k$ as its own  Lie algebra.
As another  application, consider $X$ and $P$ defined on ${\cal S}(\mathbb R)$ as usual. The three symmetric operators $I,X,P$ enjoy the Lie algebra of Weyl-Heisenberg Lie group. Moreover $\Delta = X^2+P^2 +I^2$ is essentially self adjoint on ${\cal S}(\mathbb R)$, because it admits a dense set of analytic vectors (the finite linear combinations of eigenstates of the standard harmonic oscillator). Thus these operators admit unique self-adjoint extensions and are generators of a unitary representation of the (simply connected)  Weyl-Heisenberg Lie group. This example holds also replacing $L^2$ with another generic Hilbert space $\cal H$ and $X,P$ with operators verifying CCR on an dense invariant domain where $X^2+P^2$ (and thus also $X^2+P^2 +I^2$) is essentially self adjoint. It is possible to prove that the existence of the unitary rep of the Weyl-Heisenberg Lie group, if the space is irreducible, establishes the existence of a unitary operator from ${\cal H}$ to $L^2$ transforming  $X$ and $P$ into the standard operators. Following this way one builds up an alternate  proof of Stone-von Neumann's theorem.
As a last comment, I stress that usually ${\cal G}_\ell$ is not the group acting in the physical space and this fact may create some problem: Think of $SO(3)$ that is the group of rotations  one would like to represent at quantum level, while he/she ends up with a unitary representation of $SU(2) \neq SO(3)$. Usually nothing too terrible arises this way, since the only consequence is  the appearance of annoying phases as explained by Josh, and overall phases do not affect  states. Nevertheless sometimes some disaster takes place: For instance, a physical system cannot assume  quantum states that are coherent superpositions of both integer and semi-integer spin. Otherwise an internal phase would take place after a $2\pi$ rotation. What is done in these cases is just to forbid these unfortunate superpositions. This is one of the possible ways to realize superselection rules.      
A: I'd like to add to Josh's answer, because he didn't really explain what a universal covering group is. Essentially, a space $T$ is a covering space of another space $U$ if, for an open subset of $U$, there's a function $f$ that maps a union of disjoint open subsets of $T$ to the subset of $U$. Or, more simply worded, pick a piece of your space $U$, and I'll find you some number of different pieces of $T$ that have a function mapping them onto the piece of $U$. In the case of $T = Spin(3)$ and $U = SO(3)$, there are two disjoint subsets of $Spin(3)$ for every subset of $SO(3)$, so we say that $Spin(3) \cong SU(2)$ is the double cover of $SO(3)$.
Now, the way that this relates to bosons and fermions is where Josh's answer comes in. We want physical states to live in vector spaces that carry (projective) representations of our symmetry groups. The "projective" part means that our states may pick up a phase when transformed to other states -- so, for example, if you rotate a spin-1/2 state 360$^{\circ}$, the state picks up a minus sign. It turns out that, at least in the case of $SO(3)$, we can eliminate the need for the "projective" part of this -- and thus those pesky minus signs -- by considering instead representations of the covering space.
