To understand a tensor operator one first needs to be reminded that it's an operator. Whereas an (angular momentum) state $\vert \ell m\rangle$ transform as a column vector under rotation:
\begin{align}
\vert \ell m\rangle \to R\vert \ell m\rangle &= \sum_{m'}
\vert\ell m'\rangle \langle \ell m'\vert R\vert \ell m\rangle \\
&=\vert\ell m'\rangle\, D^\ell_{m'm}(R) \tag{1}
\end{align}
an operator is a matrix so heuristically one needs one transformation $R$ for the rows and another $R^{-1}$ for the columns so that the component $m$ of the (angular momentum, or spherical) tensor operator $\ell$ transforms like
\begin{align}
T^\ell_m \to RT^\ell_m R^{-1} &= \sum_{m'}
T^{\ell}_{m'}\, D^\ell_{m'm}(R)\, . \tag{2}
\end{align}
Basically, Eq.(2) is taken to be the definition of a tensor operator. It is designed to that the component $m$ of the tensor $\ell$ transform
under rotation exactly as the state $\vert \ell m$ does under rotation.
Alternatively, for infinitesimal transformation we have
\begin{align}
\hat L_\pm \vert \ell m\rangle = \sqrt{(\ell\mp m)(\ell \pm m+1)}
\vert \ell m\pm 1\rangle \quad &\rightarrow\quad [\hat L_\pm, T^\ell_m]
= \sqrt{(\ell\mp m)(\ell \pm m+1)} T^{\ell}_{m\pm 1}\, ,\\
\hat L_z \vert \ell m\rangle = m
\vert \ell m \rangle \quad &\rightarrow\quad [\hat L_z, T^\ell_m]
= m T^{\ell}_{m }\, ,\\
\end{align}
Another way to think about this is to realize that a tensor operator acting on states $\vert s \mu\rangle$ will transform this state to another state $\vert s \mu' \rangle$ and so can be written as the linear combination
$$
T^{\ell}_{m}=\sqrt{\frac{2L+1}{2s+1}}\sum_{\mu\mu'}C^{s\mu'}_{S\mu;\ell m}
\vert s\mu'\rangle \langle s\mu\vert\, ,
$$
where $C^{s\mu'}_{S\mu;\ell m}$ is a Clebsch-Gordan coefficient. This way of writing the tensor components makes it clear that, if
$\vert \ell m\rangle \to R\vert \ell m\rangle$, then
$\langle \ell m' \vert \to \langle \ell m' \vert R^{-1}$
so that, under rotation, $T^\ell_m$ does go to $RT R^{-1}$.
Under this rotation, we now have
\begin{align}
R T^{\ell}_{m}R^{-1}&=\sqrt{\frac{2L+1}{2s+1}}\sum_{\mu\mu'}C^{s\mu'}_{S\mu;\ell m}
R \vert s\mu'\rangle \langle s\mu\vert\,R^{-1}\, ,\\
&=\sqrt{\frac{2L+1}{2s+1}}\sum_{\mu\mu'}\sum_{MM'}C^{s\mu'}_{S\mu;\ell m}
\vert sM'\rangle \langle sM'\vert R \vert s\mu'\rangle \langle s\mu\vert\,R^{-1}\vert s M\rangle\langle s M\vert\, ,\\
&=\sqrt{\frac{2L+1}{2s+1}}\sum_{\mu\mu'}\sum_{MM'}C^{s\mu'}_{S\mu;\ell m}
D^s_{M'\mu'}(R) D^s_{\mu M}(R^{-1}) \vert s M'\rangle\langle s M\vert\, .
\end{align}
With some dexterity the sum of products $\sum_{\mu\mu'}D^s_{M'\mu'}(R) D^s_{\mu M}(R^{-1})$ can be combined using the Clebsch $C^{s\mu'}_{S\mu;\ell m}$ to give a sum of type
$$
\sum_{m'}D^\ell_{m'm}(R)C^{s\mu'}_{S\mu;\ell m'}
$$
from which Eq.(2) follows.
Broadly speaking, irreducible under a set transformation means there is no subset that transforms only amongst itself under this set of transformations.
(The combination rules for $D$'s can be fond in Varshalovich, D. A., Moskalev, A. N., & Khersonskii, V. K. M. (1988). Quantum theory of angular momentum. World Scientific.)
