I will write
$$
\vert\alpha,\ell,s;j,m_j\rangle=\sum_{m_\ell m_s} C^{jm_j}_{\ell m_\ell; s m_s}
\vert \alpha \ell m_\ell\rangle \vert \alpha s m_s\rangle\, , \tag{1}
$$
and will assume that the tensor $T^{(k)}_q$ acts only on the orbital part, i.e. on the kets $\vert \alpha \ell m_\ell\rangle $. An expansion similar to (1) can be done for the
bra $\langle \alpha',\ell',s';j',m'_j\vert $ and the Wigner Eckart theorem will produce a
sum of the type
\begin{align}
\langle \alpha';\ell',s';j'm'_j\vert T^{(k)}_q \vert\alpha,\ell,s;j,m_j\rangle
&=\sum_{m_\ell m_s m_{\ell'}}
C^{jm_j}_{\ell m_\ell; s m_s} C^{j'm'_j}_{\ell' m'_\ell; s m_s}\delta_{ss'}\delta_{m_sm'_s}
\frac{\langle \alpha' \ell'\Vert T^{(k)}\Vert \alpha \ell\rangle}{\sqrt{2\ell'+1}}
C^{\ell' m_{\ell'}}_{kq;\ell m_\ell}\, \\
&=\frac{\langle \alpha j'\Vert T^{(k)}\Vert \alpha j\rangle}{\sqrt{2j'+1}}C^{j'm'}_{kq;jm}
\end{align}
The triple sum of Clebsch's on the right is actually proportional to the product of $ C^{\ell' m_{\ell'}}_{kq;\ell m_\ell}$ and a $6j$ symbol. Alternatively, one can multiply both sides by $C^{j''m''}_{kq;jm}$ and sum over $j'',m''$ to obtain a quadruple product of Clebsch's, proportional to a single $6j$ symbol. These summations can be found in
Varshalovich, Dmitriĭ Aleksandrovich, Anatolï Nikolaevitch Moskalev, and Valerii Kel'manovich Khersonskii. Quantum theory of angular momentum. 1988.
Either way, once these operations are done (they will involve permuting indices in the CGs ) one finally obtains
\begin{align}
\langle \alpha ' j' \ell' s'\Vert T^{(k)}\Vert \alpha j \ell s\rangle &=\delta_{ss'}
\sqrt{\frac{2j'+1}{2\ell'+1}}U(s \ell j' k; j \ell') \langle \alpha' \ell'\Vert T^{(k)}\Vert \alpha \ell\rangle
\tag{2}\\
&= \delta_{ss'}(-1)^{s+\ell+j'+k}
\sqrt{(2j '+1)(2j+1)}
\left\{\begin{array}{ccc}s&\ell&j\\ \ell &j&k \end{array}\right\}\langle \alpha' \ell'\Vert T^{(k)}\Vert \alpha \ell\rangle\, ,
\end{align}
where $\left\{\begin{array}{ccc}s&\ell&j\\ \ell &j&k \end{array}\right\}$ is a $6j$ symbol. There are several
version of Eq.(2), based on symmetries of the $6j$ symbols. The version given here is from
Rowe, David J., and John L. Wood. Fundamentals of nuclear models: foundational models. World Scientific Publishing Company, 2010.
Various authors use various symbols such as the $W$ or $U$ symbols of Wigner and Racah respectively.
Thank you for your clarification on $\alpha$, so I can provide the answer more accurately.
The trick to deriving the last equation is to consider the product $\langle e^{l}_{\lambda}|O^{\mu}_i|e^{\nu}_{j}\rangle$ with
\begin{align}
\langle e^{l}_{\lambda}|O^{\mu}_i|e^{\nu}_{j}\rangle & = \langle e^{l}_{\lambda}|U(g^{-1})U(g)O^{\mu}_i|e^{\nu}_{j}\rangle \\
& = \langle e^{\color{red}{k}}_{\lambda}|(D^{\lambda}(g)^{\color{red}{k}}_{l})^*\sum_{\alpha,\lambda',l'}{\bigg(D^{\lambda'}(g)^{\color{red}{q}}_{l'}|\omega^{\lambda'}_{\alpha,{\color{red}{q}}}\rangle \langle \alpha,\lambda',l'(\mu,\nu)i,j \rangle \bigg)} \\
& = \sum_{\alpha,\lambda',l'}{\langle \alpha,\lambda',l'(\mu,\nu)i,j \rangle \bigg((D^{\lambda}(g)^{\color{red}{k}}_l)^*D^{\lambda'}(g)^{\color{red}{q}}_{l'}\bigg)\langle e^{\color{red}{k}}_{\lambda}|\omega^{\lambda'}_{\alpha,{\color{red}{q}}}\rangle}
\end{align}
where red labels are summed over as Einstein's convention. Now we integrate both sides over $g$, and use the orthogonality relation to come up with
\begin{align}
\langle e^{l}_{\lambda}|O^{\mu}_i|e^{\nu}_{j}\rangle & = \sum_{\alpha,\lambda',l'}{\langle \alpha,\lambda',l'(\mu,\nu)i,j \rangle{1 \over 2\lambda+1}\delta_{\lambda\lambda'}\delta_{{\color{red}{k}}{\color{red}{q}}}\delta_{ll'}\langle e^{\color{red}{k}}_{\lambda}|\omega^{\lambda'}_{\alpha,{\color{red}{q}}}\rangle} \\
& = \sum_{\alpha}{\langle \alpha,\lambda,l(\mu,\nu)i,j \rangle{1 \over 2\lambda+1}\sum_{k}{\langle e^{k}_{\lambda}|\omega^{\lambda}_{\alpha,k}\rangle}}
\end{align}
This answers your first question, and about the second one, your thought is correct, the set of vectors $\big\{|\omega^{\lambda}_{\alpha,k} \rangle\big|\lambda=|\mu-\nu|,\cdots,\mu+\nu, \ \text{and} \ |k| \leq \lambda\big\}$ are constructed from $\big\{O^{\mu}_i|e^{\nu}_{j}\rangle\big||i| \leq \mu, \ \text{and} \ |j| \leq \nu\big\}$. Therefore, the dependence of $\mu$ and $\nu$ are not absent from the reduced matrix.
Best Answer
The key point is that $$ T^{1}_{m_1}\vert \ell_2,m_2\rangle \tag{1} $$ transforms in exactly the same way as \begin{align} \vert 1,m_1\rangle\vert \ell_2,m_2\rangle \, .\tag{2} \end{align} You can see this because \begin{align} L_zT^{1}_{m_1}\vert \ell_2m_2\rangle &= [L_z,T^{1}_{m_1}]\vert \ell 2 m_2\rangle + T^{1}_{m_1}L_z\vert \ell_2 m_2\rangle\, ,\\ &=(m_1+m_2)T^1_{m_1}\vert\ell_2 m_2\rangle \end{align} and likewise the actions \begin{align} L_\pm T^{1}_{m_1}\vert \ell_2 m_2\rangle &=[L_\pm,T^1_{m_1}]\vert \ell_2m_2\rangle + T^{1}_{m_1} L_\pm \vert \ell_2m_2\rangle\, ,\\ &\sim L_\pm \left[\vert 1m_1\rangle \vert \ell_2 m_2\rangle\right] =\left[L_\pm\vert 1m_1\rangle\right]\vert\ell_2 m_2\rangle +\vert 1m_1\rangle\left[L_\pm \vert \ell_2m_2\rangle\right] \end{align} where by $\sim$ I mean the same factors with square roots come out, and of course $[L_\pm,T^{1}_{m_1}]\propto T^1_{m\pm 1}$.
If you believe that (1) does indeed transform in the same way as (2) does, then it is clear that, if (2) yields states with $j$ values ranging from $\ell_2+1$ to $\vert \ell_2-1\vert$, then the combination in (1) will also yields states with values of $j$ in that range. That’s basically the selection rule $\Delta j=0,\pm 1$.
Obviously if you do this not with a vector operator but a general tensor $T^{\ell_1}$, you will find the range of $j$ is just the same as the range in $\ell_1\otimes\ell_2$.