Let $\mathcal{A},\mathcal{B}$ be small categories, $\mathcal{C}$ a cocomplete category and $\mathcal{D}$ an arbitrary category. Consider functors $F:\mathcal{A}\rightarrow\mathcal{B}$, $G:\mathcal{A}\rightarrow\mathcal{C}$, $R:\mathcal{D}\rightarrow\mathcal{C}$ and $L:\mathcal{C}\rightarrow\mathcal{D}$, where $L$ is left adjoint to $R$. We want to show that
$$L\circ\text{Lan}_F(G)=\text{Lan}_F(L\circ G).$$
The author provides the following proof:
For every functor $H:\mathcal{B}\rightarrow\mathcal{D}$, we get the
following bijections:$$ \begin{align} \text{Nat}\left(L\circ\text{Lan}_F(G),H\right) &
\cong \text{Nat}\left(\text{Lan}_F(G),R\circ H\right) \\ & \cong
\text{Nat}\left(G,R\circ H\circ F\right) \\ & \cong
\text{Nat}\left(L\circ G,H\circ F\right)\\ & \cong
\text{Nat}\left(\text{Lan}_F(L\circ G),H\right) .\end{align}$$
What is the justification for these isomorphisms; i.e. how does one deduce them? On the other hand, are these classes of natural transformations sets?
Edit:
Let $R_*:[\mathcal{B},\mathcal{D}]\rightarrow[\mathcal{B},\mathcal{C}]$ be the functor defined by $R_*(H)=R\circ H$ and $R_*(\alpha)=R*\alpha$. Then $R_*$ is left adjoint to $L_*$. But I am not sure how this helps. If $\mu:L\circ\text{Lan}_F(G)\Rightarrow H$, then
$$R_*(\mu):R\circ L\circ\text{Lan}_F(G)\Rightarrow R\circ H$$
–but this doesn't give a natural transformation from $\text{Lan}_F(G)$ to $R\circ H$.
Best Answer
If $L\dashv R$, then $L_*\dashv R_*$, not the other way around (the places are switched if you take precomposition, $L^*, R^*$)
A good way to see that $L_*\dashv R_*$ is via the formulation of adjoints with a unit and a counit satisfying the triangle identities.
You'll want $\epsilon_* : L_*R_*\to id$, which is simply given by $\epsilon G : LRG\to G$ for any $G$, and $\eta_* : id\to R_*L_*$ which is also given by $\eta G : G\to RLG$. That they satisfy the triangle identities is essentially obvious because $\epsilon,\eta$ do (if you're not convinced, write it down !)
Therefore all the isomorphisms in the proof are justified.
(with regards to your edit, maybe a more concrete way to see it : suppose you have $\theta : LT\to S$, then you get $R\theta : RLT\to RS$, and then you can precompose by $\eta T : T\to RLT$ to get $(R\theta)\circ \eta T : T\to RS$)