I am trying to understand a concept of Kan extension. It has a description in terms of limits. I want to see what happens in an example.
Let $[0]$ be a category with one object $a$ and one morphism $\text{Id}_a$.
Let $G$ be a group. Denote by $BG$ a category with single object $b$ and $Hom_{ \mathcal{G} } (b, b) = G$.
Let $\mathcal{Set}$ be the category of sets.
Let $X$ be a functor from $[0]$ to $\mathcal{Set}$, sending $a$ to $x \in \mathcal{Ob}( \mathcal{Set} )$.
Let $F$ be the unique functor from $[0]$ to $BG$
Finally let $R = Ran_F (X)$ is the right Kan extension of $X$ along $F$.
\begin{array}{111} [0] & \overset{X}{\longrightarrow} & \mathcal{Set}\\ \Big\downarrow{F} & \nearrow{R} \\ BG \\ \end{array}
The description suggest that one should take the limit over the comma category. Comma category is the category with $\#G$ objects, and all maps are identity. The limit of $X(a) = x$ over this category is $x^{ \#G}$.
The definition of right Kan extension suggests that there exist $\eta: RF \rightarrow X$. It means that there is a map $\eta_a : x^{ \#G } \rightarrow x$.
Question 1 What is the map $\eta_a$?
Question 2 How to prove that the description valid in this example by direct elementary argument?
The natural transformation is just one of the product projections from $F(a)^{|G|}$; in particular, it's the projection corresponding to $(a,id_{F(a)})$ in the comma category $(F(a)\downarrow F)$. Which in this case is really just the function $f\mapsto f(e)$, where $e$ is the unit of the group. It's not too much work to see why this turns out to be natural in this particular case.
To see why it turns out to be natural in the general case, we'll assume $\mathbf{A},\mathbf{B},\mathbf{C}$ and $X,F$ as in the Wikipedia article you link to above, letting $\{s^b_{(a,f)}\}_{(a,f)\in(b\downarrow F)}$ be the limiting cone for the limit we take as $R(b)$, then $\eta_a:RF(a)\to X(a)$ is the component $s^{F(a)}_{(a,id_{F(a)})}$. Naturality works here because for $g:a\to a'$ in $\mathbf{A}$, $RF(g)$, by the limit construction, is the unique morphism $\alpha$ such that $s^{F(a')}_{(a'',f)}\circ\alpha=s^{F(a)}_{(a'',F(g)\circ f)}:RF(a)\to X(a')$. In the special case of $(a',id_{F(a')})$, $$s^{F(a')}_{(a',id_{F(a')})}\circ RF(g)=s^{F(a)}_{(a,F(g))}=X(g)\circ s^{F(a)}_{(a,id_{F(a)})}$$. This last part holds because $g$ is a morphism $(a,id_{F(a)})\to (a',F(g))$ in $(F(a)\downarrow F)$ and $s^{F(a)}$ is natural.
For why it has the universal property intended, I recommend either reading the chapter in Mac Lane's Categories for the Working Mathematician or the one in Borceux's Handbook of Categorical Algebra. It's not hard to show, but it's a similar degree of fidgety to the proof of naturality above.