Since two dimensions is enough to faithfully represent $SU(2)$, how can there be irreducible representations of dimension $n>2$? I am new to Lie goups and representation theory, and I am confused about irreducible representations of $SU(2)$ of dimension $n>2$.
Let us take the more intuitive example of $SO(2)$. I seems obvious that any faithful 3-dimensional representation is similar to a representation of the form
$$\begin{pmatrix}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{pmatrix}.$$
This representation is clearly the direct sum of a 2-dimensional representation of $SO(2)$ and the trivial representation, so any 3-dimensional representation of $SO(2)$ is reducible. Is this basically correct?
It seems to me like the same should be true for $SU(2)$, but I am reading that spin-1 particles transform according to a 3-dimensional representation, that spin-3/2 particles transform according to a 4-dimensional representation; and so on. In fact, Wikipedia states that "There is one n-dimensional irreducible representation of SU(2) for each dimension".
How can this be? I.e., how can there be irreducible representations of $SU(2)$ of dimension $n>2$, when two dimensions is "enough"?
 A: Recall that a representation $\Pi: G \mapsto \mathrm{GL}(V)$ is irreducible if there are no non-trivial subspaces $U\subsetneq V$ such that $\Pi(G) U = U$.  Your intuition appears to be that given a Lie group $G$, there is some positive integer $n$ such that representations on spaces with dimension $\gt n$ must be reducible, and representations on spaces with dimension $\lt n$ must be unfaithful.  This isn't right.
If $G$ is $n$-dimensional as a manifold, then its image $\Pi(G)$ under a faithful representation must be an $n$-dimensional submanifold of $\mathrm{GL}(V)$.  This certainly sets a lower limit on the dimensionality of $V$, e.g. $\mathrm{SO}(3)$ is a 3-dimensional manifold which cannot be faithfully represented on $\mathbb R$, as $\mathrm{GL}(\mathbb R)\simeq (\mathbb R_{\neq 0},\cdot)$ is a 1-dimensional manifold.
However, there is no a priori upper limit on the dimensionality of $V$.  A comparatively low-dimensional $\Pi(G)$ may still mix $V$ together in such a way that no non-trivial subspaces $U\subsetneq V$ are left invariant.  For example, consider the following representations of $S^1$ (which we will define as $\mathbb R\ \mathrm{mod}\ 2\pi$):
$$\Pi_1 :S^1 \rightarrow GL(\mathbb C), \qquad \theta \mapsto e^{i\theta}$$
$$\Pi_2 : S^1 \rightarrow GL(\mathbb R^2), \qquad \theta \mapsto \pmatrix{\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) }$$
$\Pi_1$ is a representation on the one-dimensional vector space $\mathbb C$, while $\Pi_2$ is a representation on the two-dimensional vector space $\mathbb R^2$.  Both representations are faithful and irreducible, contradicting the notion that the reducibility of a representation can be inferred from the dimensionality of the representation space.
This example isn't ideal, in the sense that it's not immediately clear that the apparent decoupling between irreducibility and dimensionality isn't somehow due to the fact that the first representation is complex while the second is real.  In fact it is not - as you state in the question, there exist faithful, irreducible representations of $\mathrm{SU}(2)$ for all $\mathbb C^n$ with $n>1$.  These can be obtained by exponentiating the higher-dimensional analogues of the Pauli matrices, but doing this explicitly is an algebraic nightmare.
A: Consider the simple case where there are two spin-1/2 particles.  The possible states of such a system are
\begin{align}
\vert ++\rangle\, ,\qquad 
\vert +-\rangle\, ,\qquad 
\vert -+ \rangle\, ,\qquad 
\vert --\rangle
\end{align}
and the total angular momentum operators are
\begin{align}
\hat L_{i}= {\hat L}_i^{(1)}
+{\hat L}_i^{(2)}
\end{align}
Construct the total raising and lowering operators in the usual way and clearly you will find that the $3$ states
\begin{align}
\vert ++\rangle\, ,\qquad 
\frac{1}{\sqrt{2}}\left(\vert +-\rangle + \vert -+\rangle \right)\, ,
\qquad 
\vert --\rangle \tag{1}
\end{align}
transform amongst themselves under the angular momentum operators.  They span a basis for a 3-dimensional representation.
Thus, you need this 3-dimensional representation (with total spin $S=1$) to describe two-spin systems.  You also need the $S=0$ state
$$
\frac{1}{\sqrt{2}}\left(\vert +-\rangle - \vert -+\rangle \right)
$$
but this state will never connect to the three states in (1) so it lives in a different subspace of the original 4-dimensional space.
One can consider additional examples:  the inertia tensor is constructed from quantities of the type $x_i x_j$, so that if
$\{x_i,i=x,y,z\}$ transform by a 3-dimensional representation of SO(3), then one can show that the pieces of inertia tensor (which is symmetric in its entries) transform by a 5-dimensional ($L=2$) representation of SO(3), and another 1-dimensional $(L=1)$ piece.
One can think of $SU(2)$ irreps like $S=1, 3/2$ as occurring when combining multiple $s=1/2$ particles.  That's one way they will occur.   It is true that one can infer their properties under $SU(2)$ transformation by looking at the transformations of the individual constituents, but that's not necessarily practical (you'd need to construct the basis states in terms of $n$ constituents all the time) and it's not mathematically necessary either.  What matters (at the algebra level) is that expression of the angular momentum operators in the space spanned by the $2S+1$ states of spin $S$ satisfy the same commutation relations as those in the 2-dimensional space: that's the core definition of a representation.
Although it may be convenient to think of higher $S$ as coming from putting together many particles with $s=1/2$, the representations exists "on their own" without reference to multi particle systems.  It is important to keep in mind that the mathematics here is descriptive, i.e. we use it because it's convenient and it allows us to correctly model what we see.  In other words, we have use for these higher $S$ irreps without necessarily thinking of the basis states as composites of spin-1/2 elements.
Now, it turns out that irreps of SO(2) are actually 1-dimensional (over the complex) because SO(2) is abelian, so in fact your statement is incorrect.  Using complex coordinates, you can rewrite your element as
\begin{align}
\hbox{diag}\left(e^{i\theta} , e^{-i\theta},1\right)
\end{align}
so in fact your example breaks up into a direct sum of 3 irreps.  In general, the irreps of SO(2) are labelled by $k$ so that
$g(\theta)\in SO(2)\mapsto e^{ik\theta}$ in irrep $k$.
