The statement in the title of the question is in my lecture notes. I don't understand the reason for this.
Furthermore, I'm having trouble understanding what spin is in the first place. I know that for a fermion, which has spin 1/2, the wavefunction picks up a phase of $\pi$ under an "internal rotation" of $2 \pi$. Does this have anything to do with the above statement?
The Hilbert space for a single particle is $\mathcal H= L^2(\mathbb R^3)\otimes \mathcal H_{\text{internal}}$, where $\mathcal H_{\text{internal}}$ is a Hilbert space which describes the internal degrees of freedom. On $\mathcal H$, we need a unitary action of the group of isometries of $\mathbb R^3$, namely the semi-direct product $\mathbb R^3 \rtimes SO(3)$.
Isometries, angular momentum and spin
The translation group $\mathbb R^3$ acts in the usual way on $L^2(\mathbb R^3)$ and trivially on $\mathcal H_{\text{internal}}$. Physically, this means that the particle doesn't carry internal momentum, because the generators of translation are the momentum operators.
The rotation group acts in the usual way on $L^2(\mathbb R^3)$. It's action on $\mathcal H_{\text{internal}}$ is what defines spin. Physically, this means that the particle can carry some angular momentum internally, because the generators of rotation are angular momentum operators (see below). Now, because absolute phase isn't physical in quantum mechanics, we can allow the action to only be projective, ie $\mathcal H_{\text{internal}}$ is a representation of $SU(2)$.
We can also require this representation to be irreducible. (Mathematically, in non-relativistic QM, it is possible to have a particle which is allowed to be in a superposition of spin $0$ and spin $1/2$.)
Those irreducible representation are indexed by some number $s$ (integer or half-integer), in which case the dimension is $2s+1$. This is what we call spin.
Total angular momentum
For a rotation $R$, the unitary operator representing this symmetry on $\mathcal H$ is $U(R) = U_{L^2(\mathbb R^3)}(R)\otimes U_s (R)$ where the first factor is the usual rotation operator on wave functions and $U_s(R)$ is in the representation of spin $s$.
For infinitesimal $Rx \simeq x +\epsilon y\times x$, we have, by definition $U(R) = \mathbb I + i\epsilon y_k J_k$ where $J_k$ are the total angular momentum operators. In our case, we have : \begin{align} U_{L^2(\mathbb R^3)}(R) &\simeq \mathbb I + i\epsilon y_kL_k \\ U_s(R) &\simeq \mathbb I + i\epsilon y_k S_k \end{align} where $L_k$ are the usual orbital angular momentum operators and $S_k$ are the spin operators.
Matching both sides of $U(R) = U_{L^2(\mathbb R^3)}(R)\otimes U_s (R)$, we see that : $$J_k = L_k + S_k$$
In other words, the internal degrees of freedom of the particle allow it to carry some angular momentum.
Multi-component wave-functions $\newcommand{\ket}[1]{\left|#1\right\rangle} \newcommand{\bra}[1]{\left\langle #1\right|} \newcommand{\braket}[2]{\left\langle #1 \middle|#2\right\rangle}\newcommand{\mel}[3]{\left\langle #1 \middle| #2 \middle|#3 \right\rangle}$
The position eigenstates $\ket x$ form a basis of $L^2(\mathbb R^3)$. For a general state $\ket \psi$, the wave function is defined as $\psi(x) = \braket{x}{\psi}$.
We take a basis $\ket k$ of the internal spin Hilbert space, where $k$ ranges from $-s$ to $s$ in integer steps, taking on $2s+1$ values. Assuming that there are no other internal degrees of freedom, a basis for the full Hilbert space is given by the $\ket x \otimes \ket k$. Given a state $\ket \psi$, we can therefore define $2s+1$ wave-functions (or a wave-function with $2s+1$ components) by :
$$\psi_k(x) = \left( \bra x \otimes \bra k\right) \ket \psi$$
Phase under a rotation of $2\pi$
The integer spin representation are single-valued : after a rotation of $2\pi$, the state is brought back to its initial value. For half-integer spin, the state picks up a minus sign after a rotation of $2\pi$. Given what we saw above, the same goes for the different components of the wave-function.
