What is the physical significance of the double connectivity of $\rm SO(3)$ group manifold? Is there any physical significance of the fact that the group manifold (parameter space) of $SO(3)$ is doubly connected?
There exists two equivalence classes of paths in the group manifold of SO(3) or in other words, $\Pi_1(SO(3))=\mathbb{Z}_2$. This space is, therefore, doubly connected. There are paths which come back to initial configurations after a rotation of $2\pi$ and others after a rotation of $4\pi$, with proper parametrization of angles.
Using this fact, is it possible to show that such a topology admits the existence of half-integer spins and integer spins? I understand spinors as objects whose wavefunctions pick up a negative sign after a rotation of $2\pi$, and comes back to itself after a rotation of $4\pi$. Right? But from the topological argument given above, it is not clear to me, that how does it lead to two kinds of  wavefunctions, spinor-type $(j=\frac{1}{2},\frac{3}{2},\frac{5}{2}...)$ and tensor-type $j=0,1,2,...$? 
It is not explicitly clear how these two types of paths in SO(3) group manifold will lead to such transformation properties on "the wavefunctions"?
http://en.wikipedia.org/wiki/SO%283%29#Topology. 
 A: Just in view of the double universal covering provided by $SU(2)$,  $SO(3)$ must  a quotient of $SU(2)$ with respect to a central discrete normal subgroup with two elements. This is consequence of a general property of universal covering Lie groups: 
If $\pi: \tilde{G} \to G$ is the universal covering Lie-group homomorphism, the kernel $H$ of $\pi$ is a discrete normal central subgroup of the universal covering $\tilde{G}$ of $G= \tilde{G}/H$, and $H$ is isomorphic to the fundamental group of $G$, i.e. $\pi_1(G)$ (wich, for Lie groups, is Abelian) .
One element of that subgroup must be $I$ (since a group includes the neutral element). The other, $J$, must  verify  $JJ=I$ and thus $J=J^{-1}= J^\dagger$. By direct inspection one sees that in $SU(2)$ it is only possible for $J= -I$. So   $SO(3) = SU(2)/\{I,-I\}$. 
Notice that $\{I,-I\} = \{e^{i4\pi \vec{n}\cdot \vec{\sigma}/2 }, e^{i2\pi \vec{n}\cdot \vec{\sigma}/2 }\}$ stays in the center of $SU(2)$, namely the elements of this subgroup commute with all of the elements of $SU(2)$. Moreover $\{I,-I\}=: \mathbb Z_2$ is just the first homotopy group of $SO(3)$ as it must be in view of the general statement I quoted above. 
A unitary representations of $SO(3)$ is also a representation of $SU(2)$ through the projection Lie group homomorphism $\pi: SU(2) \to SU(2)/\{I,-I\} = SO(3)$. So, studying unitary reps of $SU(2)$ covers the whole class of unitary reps of $SO(3)$.  Let us study those reps.
Consider a unitary representation $U$ of $SU(2)$ in the Hilbert space $H$. The central subgroup $\{I,-I\}$ must be represented by $U(I)= I_H$ and $U(-I)= J_H$, but $J_HJ_H= I_H$ so, as before, $J_H= J_H^{-1}= J_H^\dagger$. 
As $J_H$ is unitary and self-adjoint simultaneously, its spectrum has to be included in $\mathbb R \cap \{\lambda \in \mathbb C \:|\: |\lambda|=1\}$. So (a) it is made of $\pm 1$ at most and (b) the spectrum is a pure point spectrum and so only proper eigenspeces arise in its spectral decomposition.
If $-1$ is not present in the spectrum, the only eigenvalue is $1$ and thus $U(-I)= I_H$. If only the eigenvalue $-1$ is present, instead, $U(-I)= -I_H$. 
If the representation is irreducible $\pm 1$ cannot be simultaneously eigenvalues. Otherwise $H$ would be split into the orthogonal direct sum of eigenspaces $H_{+1}\oplus H_{-1}$. As $U(-1)=J_H$ commutes with all $U(g)$ (because $-I$ is in the center of $SU(2)$ and $U$ is a representation), $H_{+1}$ and $H_{-1}$ would be invariant subspaces for all the representation and it is forbidden as $U$ is irreducible. 
We conclude that, 
if $U$ is an irreducible unitary representation of $SU(2)$, the discrete normal subgroup $\{I,-I\}$ can only be represented by either $\{I_H\}$ or $\{I_H, -I_H\}$.
Moreover:
Since $SO(3) = SU(2)/\{I,-I\}$, in the former case $U$ is also a representation of $SO(3)$. It means that $I = e^{i  4\pi \vec{n}\cdot \vec{\sigma} }$ and $e^{i  2\pi \vec{n}\cdot \vec{\sigma}/2 } = -I$ are both transformed into $I_H$ by $U$.
In the latter case, instead,  $U$ is not a true representation of $SO(3)$, just in view of a sign appearing after $2\pi$, because $e^{i  2\pi \vec{n}\cdot \vec{\sigma}/2 } = -I$ is transformed into $-I_H$ and only     $I = e^{i  4\pi \vec{n}\cdot \vec{\sigma}/2 }$ is transformed into $I$ by $U$.
