My question is a very basic one. In physics, we talk about the
$$ \left( \frac{1}{2}, \frac{1}{2} \right) \qquad \text{and} \qquad \left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) $$
of SU(2)$\times$SU(2) (I think). What exactly does this notation mean? I see it everywhere but nobody ever explains it. Is the former just saying that each of the SU(2)s is in the fundamental representation? Alright then, if that's true. But then what exactly does the latter mean?
In the first, pairs of matrices $(A,B)$ act on tensors $u\otimes v\in \mathbb{C}^2\otimes\mathbb{C}^2$ via
$$ (A,B)(u\otimes v)=Au\otimes Bv \tag{1}$$
whereas in the second, pairs of matrices act on vectors $(u,v)\in\mathbb{C}^2\oplus\mathbb{C}^2$ via
$$ (A,B)(u,v)=(Au,Bv). \tag{2}$$
I use the $\otimes$ symbol since I do math not physics. Mathematically, in $U\oplus V$ we have
$$ (u_1,v_2)+(u_2,v_2)=(u_1+u_2,v_1+v_2) \tag{3}$$
where $u_1,u_2\in U,v_1,v_2\in V$. (This does not apply to the spin "$\frac{1}{2}$" in $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ because $\frac{1}{2}$ is not a vector in a vector space, it's just a label.) If $\{u_1,\cdots,u_m\}$ is a basis for $U$ and $\{v_1,\cdots,v_n\}$ is a basis for $V$ then $\{u_1,\cdots,u_m,v_1,\cdots,v_n\}$ is a basis for $V$. Therefore $$\dim(U\oplus V)=\dim(U)+\dim(V)\tag{4}$$
On the other hand, in $U\otimes V$ we have
$$ u_1\otimes v_1+u_2\otimes v_2 \color{Red}{\ne} (u_1+u_2)\otimes(v_1+v_2). \tag{5}$$
Instead we have
$$ (u_1+u_2)\otimes v=u_1\otimes v+u_2\otimes v \\ u\otimes(v_1+v_2)=u\otimes v_1+u\otimes v_2 \tag{6}$$
i.e. the symbol $\otimes$ behaves like multiplication with "distributive laws." If $\{u_1,\cdots,u_m\}$ is a basis for $U$ and $\{v_1,\cdots,v_n\}$ is a basis for $V$ then $\{u_i\otimes v_j\mid 1\le i\le m,1\le j\le n\}$ is a basis for $U\otimes V$. Therefore
$$ \dim(U\otimes V)=\dim(U)\times\dim(V). \tag{7}$$
It just so happens in this case that $2+2=2\times 2$, so both your reps are $4$-dimensional.
Symbolically, $(1)$ and $(2)$ look similar, but they are not the same.
Say we take the coordinate basis of the tensor product, $\{e_1\otimes e_1,e_1\otimes e_2, e_2\otimes e_1,e_2\otimes e_2\}$ so we can treat it like $\mathbb{C}^4$. And also $\{(e_1,0),(e_2,0),(0,e_1),(0,e_2)\}$ in order to treat the direct sum as $\mathbb{C}^4$ as well. Then the action of $(A,B)$ on these two $\mathbb{C}^4$s can be described as follows: let
$$ A=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}, \quad B=\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} \tag{8}$$
so for $\mathbb{C}^4$ we can write the Kronecker product
$$ A\otimes B = \begin{bmatrix} a_{11}B & a_{12}B \\ a_{21}B & a_{22}B\end{bmatrix} = \begin{bmatrix} a_{11}b_{11} & a_{11}b_{12} & a_{12}b_{11} & a_{12}b_{12} \\ a_{11}b_{21} & a_{11}b_{22} & a_{12}b_{21} & a_{12}b_{22} \\ a_{21}b_{11} & a_{21}b_{12} & a_{22}b_{11} & a_{22}b_{12} \\ a_{21}b_{21} & a_{21}b_{22} & a_{22}b_{21} & a_{22}b_{22} \end{bmatrix}. \tag{9}$$
This describes how $SU(2)\times SU(2)$ acts on $\mathbb{C}^4$ in the $(\frac{1}{2},\frac{1}{2})$ representation.
But in the $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ representation of $SU(2)\times SU(2)$ on $\mathbb{C}^4$, the pair $(A,B)$ acts by
$$ A\oplus B=\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & 0 & 0 \\ a_{21} & a_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12} \\ 0 & 0 & b_{21} & b_{22} \end{bmatrix}. \tag{10}$$
The representations are not equivalent, i.e. there's no $4\times 4$ matric $C$ such that $$C(A\otimes B)C^{-1}=A\oplus B\tag{11}$$ for all pairs $(A,B)$ in $SU(2)\times SU(2)$.
Indeed, the $(\frac{1}{2},\frac{1}{2})$ representation is irreducible (it has no invariant subspaces, although this may not be obvious directly from the description) but the $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ representation is clearly reducible (the first and second copy of $\mathbb{C}^2$ within $\mathbb{C}^4$ are invariant).