It is first important to note that it doesn't matter if you write a state in the
$| l \; m_l\; s \; m_s \rangle$ basis or the $| l \;s\; j \; m \rangle$, you'll always have that $m = m_l + m_s$.
This means that
$$| l = 2, m_l = 2,s=1/2, m_s=1/2 \rangle = | l=2, s=1/2, j=5/2, m=5/2 \rangle$$ because these are the only kets in each of the bases that have $m = m_l + m_s = 5/2$.
From that you can use the lowering operators to construct the other states.
For example, if you want to write $| l=2, s=2, j=5/2, m=3/2 \rangle$ in the $| l \; m_l\; s \; m_s \rangle$ basis, you only need to apply the $j_-=L_-+S_-$ operator to both sides of the equation.
In the left side you apply
$$
(L_-+S_- )| l \, m_l\, s \, m_s \rangle = L_-| l \, m_l\, s \, m_s \rangle+S_-| l \, m_l\, s \, m_s \rangle,
$$
with
$$ L_-| l \; m_l\; s \; m_s \rangle = \hbar\sqrt{l(l+1)-m_l(m_l-1)}| l \; m_l-1\; s \; m_s \rangle
$$
and
$$ S_-| l \; m_l\; s \; m_s \rangle = \hbar\sqrt{s(s+1)-m_s(m_s-1)}| l \; m_l\; s \; m_s-1 \rangle.
$$.
On the right side
$$
j_- | l \;s\; j \; m \rangle = \hbar\sqrt{j(j+1)-m(m-1)} | l \;s\; j \; m-1 \rangle
$$
Put everything together and you'll find what you're after.
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 $\ell_1\otimes\ell_2$.
Best Answer
When combining two angular momenta, it results from this a number of possible values of $j$'s.
In the semiclassical model, the vectors $\vec j_1$ and $\vec j_2$ can be added to produce values of $\vec j$ beyond the simple addition $\vec j_1+\vec j_2$ because the orientations of $\vec j_1$ and $\vec j_2$ need not be colinear. Quantum mechanically, this means that $\vert j_1m_1\rangle \vert j_2m_2\rangle$ will not in general be eigenstates of the total operator $\hat J^2= (\hat J_1+\hat J_2)^2$ as eigenstates of $\hat J^2$ have $j$ as a "good" quantum number.
The values of the projection on the $\hat z$ axes, however, are scalars and additive, i.e. $m_1+m_2=m$ for any of the possible values of $\vec j$.
The Clebsch-Gordan coefficient $\langle jm\vert j_1m_1;j_2m_2\rangle$ with $m=m_1+m_2$ is just the amplitude of finding the state $\vert jm\rangle$ in the coupled state $\vert j_1m_1\rangle \vert j_2m_2\rangle$, just as $\langle \phi\vert \psi\rangle$ is the amplitude of finding the state $\vert \phi\rangle$ in the state $\vert \psi\rangle$.
In particular, $\vert \langle jm\vert j_1m_1;j_2m_2\rangle\vert^2 $ is the probability of finding $\vert jm\rangle $ in $\vert j_1m_1\rangle\vert j_2m_2\rangle$. This implies, for instance, the normalization condition $$ \sum_j\vert \langle jm\vert j_1m_1;j_2m_2\rangle\vert^2=1 $$
Alternatively, to obtain the state $\vert jm\rangle$ for specificed $j$, one must consider linear combinations of states $\vert j_1m_1\rangle\vert j_2m_2\rangle$ with different $m_1$, $m_2$ but such that $m_1+m_2=m$. The coefficients of the linear combinations are precisely the CGs. Thus $$ \vert jm\rangle =\sum_{m_1m_2} \vert j_1m_1\rangle \vert j_2m_2\rangle \langle j_1m_1;j_2m_2\vert jm\rangle \tag{1} $$ In this interpretation, the probability amplitude of finding $\vert j_1m_1\rangle \vert j_2m_2\rangle$ in the state $\vert jm\rangle$ is the CG, and it thus follows that $$ \sum_{m_1m_2}\vert \langle j_1m_1;j_2m_2\vert jm\rangle\vert^2=1\, . \tag{2} $$ In (1) and (2), the sum over $m_1$ restricted to that $m_1+m_2=m$. Finally, note that since the CGs are real, we have $$ \langle j_1m_1;j_2m_2\vert jm\rangle= \langle jm \vert j_1m_1;j_2m_2\rangle\, . $$