You can decompose a rank two tensor $X_{ab}$ into three parts:
$$X_{ab} = X_{[ab]} + (1/n)\delta_{ab}\delta^{cd}X_{cd} + (X_{(ab)}-1/n \delta_{ab}\delta^{cd}X_{cd})$$
The first term is the antisymmetric part (the square brackets denote antisymmetrization). The second term is the trace, and the last term is the trace free symmetric part (the round brackets denote symmetrization). n is the dimension of the vector space.
Now under, say, a rotation $X_{ab}$ is mapped to $\hat{X}_{ab}=R_{a}^{c}R_{b}^{d}X_{cd}$ where $R$ is the rotation matrix. The important thing is that, acting on a generic $X_{ab}$, this rotation will, for example, take symmetric trace free tensors to symmetric trace free tensors etc. So the rotations aren't "mixing" up the whole space of rank 2 tensors, they're keeping certain subspaces intact.
It is in this sense that rotations acting on rank 2 tensors are reducible. It's almost like separate group actions are taking place, the antisymmetric tensors are moving around between themselves, the traceless symmetrics are doing the same. But none of these guys are getting rotated into members "of the other team".
If, however, you look at what the rotations are doing to just, say the symmetric trace free tensors, they're churning them around amongst themselves, but they're not leaving any subspace of them intact. So in this sense, the action of the rotations on the symmetric traceless rank 2 tensors is "irreducible". Ditto for the other subspaces.
When we say that $3\otimes 3 \approx 5 \oplus 3 \oplus 1$, this means that under an $SO(3)$ rotation, the first five components mix together alone "traceless symmetric tensor", so do the next 3, while the last component "the trace" is invariant.
If you prove that each of these three parts of the sum is indeed "irreducible", this means that this decomposition is UNIQUE! because in any other decomposition you are going to mix some of the "5" components with some of the "3", which would not work because these parts that you took out will definitely mix with the ones you left out (I'm trying to be as colloquial as possible).
We can also give physical arguments why a composite particle of two sub particles each having spin one (boson) cannot possibly be a half integer particle (fermion), which is what would happen if any of the representations being summed on the right hand side was even.
For the second part of your question then the unique decomposition of a tensor in $4D$ is
$4\otimes 4 \approx 9 \oplus \bar{3} \oplus 3 \oplus 1$
which are the same as before, but now the antisymmetric component splits into a dual and anti-self dual irreducible representations. (which from the point of view of the $SO(3)$ subgroup behave like a pseduo-vector and a polar vector)
REPLY TO YOUR COMMENT:
Here is where it might get confusing.. If the group itself is a product, then its representation will also be $\otimes$ but you can't convert them into sums because the final answer should always be a product. This is why you see physicists denote a representation of $SU(2)\otimes SU(2)$ by $(n,m)$.. because it ensures that whatever manipulations you do, you will end up with something of the form $(n,m)$ even though this is just $n\otimes m$ in a sense.. Now to give you some examples of how representations of $SO(4)$ are related to those of $SU(2)\otimes SU(2)/Z2$: $4 \longrightarrow (2,2)$ for a four vector which can be thought of as a product representation of two spinors, each one in different part of the $SU(2)$. An antisymmetric 4-tensor: $(4\otimes 4)_A \longrightarrow (3,1)\oplus (1,3)$ and so on
Best Answer
An order-2 tensor is usually decomposed in that way because each part behaves differently under a certain symmetry. For example, if the symmetry is just rotation, then the term with the trace transforms like a scalar; the anti-symmetric part $M_{ij}-M_{ji}$ of the tensor transforms like a pseudo-vector, while the traceless symmetric part (the last term) transforms like an ordinary 2-tensor.