$\def\Lan{\operatorname{Lan}} \def\J{\mathsf{J}} \def\C{\mathsf{C}} \def\ob{\operatorname{ob}}$I am working through Exercise 5.5.v of Riehl's Category Theory in Context. The other question previously asked here about it is this. There, OP says they have solved part (iii) on their own, without explaining how. The thing is that I am stuck while trying to do part (iii) (no answer to the question goes deeper into the solution than the point where I'm stuck).
Here's what I've done so far in each part of the exercise:
(i) $\Lan F$ sends a morphism $\alpha:j\to k$ in $\J$ to $$ \coprod_{x\in\J}\coprod_{\beta\in\J(x,j)}i_{\alpha\beta}^x: \coprod_{x\in\J}\coprod_{\beta\in\J(x,j)}Fx \longrightarrow \coprod_{x\in\J}\coprod_{\beta\in\J(x,k)}Fx, $$ where $i^y_\gamma:Fx\to\coprod_{x\in\J}\coprod_{\beta\in\J(x,j)}Fx$ is the inclusion on the index $(y,\gamma)$, for $y\in\J$, $\gamma\in\J(y,j)$.
(ii) A morphism $\nu:F\to G$ in $\C^{\ob\J}$ is just a choice of a morphism $\nu_j:Fj\to Gj$, for each $j\in\J$. Then, the component of $\Lan\nu:\Lan F\to\Lan G$ at $j\in\J$ equals $$ (\Lan\nu)_j=\coprod_{x\in\J}\coprod_{\beta\in\J(x,j)}\tilde{i}^x_{\alpha\beta}\nu_x, $$ where $\tilde{i}^y_\gamma:Gx\to\coprod_{x\in\J}\coprod_{\beta\in\J(x,j)}Gx$ is the inclusion on the index $(y,\gamma)$, for $y\in\J$, $\gamma\in\J(y,j)$.
(iii) We will verify the definition of a left adjoint via the universal arrow formulation. Given $F:\ob\J\to\C$, we define $\eta_F:F\to U\Lan F$ to have component at $j\in\J$ equal to $i^j_{1_j}$. Let $G:\J\to\C$ be any functor and $\nu:F\to UG$ be any morphism. We must verify that there is a unique $$ \tag{1}\label{nat} \bar{\nu}:\Lan F\to G $$ such that $\nu=U(\bar{\nu})\cdot\eta_F$. To see existence, define $\bar{\nu}_j=\coprod_{x\in\J}\coprod_{\beta\in\J(x,j)}G\beta\cdot\nu_x$. It is easy to see that these assemble into a natural transformation as in \eqref{nat}. It is left to see uniqueness. Suppose $\mu:\Lan F\to G$ satisfies $$ \tag{2}\label{cond} \nu=U(\mu)\cdot\eta_F $$ Write $\mu_j=\coprod_{x\in\J}\coprod_{\beta\in\J(x,j)}(\mu_j)^x_{\beta}$. Then \eqref{cond} means $(\mu_j)^j_{1_j}=\nu_j$. Naturality of $\mu$ means that for each $\alpha:j\to k$ in $\J$ we have $$ \tag{3}\label{omg} (\mu_j)^x_{\alpha\beta}=G\alpha\cdot(\mu_j)^x_{\beta}. $$ Particularizing $j=x$ and $\beta=1_j$, we deduce $(\mu_j)^j_\alpha=G\alpha\cdot\nu_j=(\bar{\nu}_j)_\alpha^j$. From here I don't know how to continue. How do we show $(\mu_j)^x_\alpha=(\bar{\nu}_j)_\alpha^x$ in general?
$\newcommand{\J}{\mathsf{J}}\newcommand{\ob}{\mathsf{Ob}}\newcommand{\lan}{\operatorname{Lan}}$Before we get into computational details, here's one way to get this "immediately": $$\mathsf{Lan}_\iota(F)\cong\int^{j\in\ob(\J)}\J(j,-)\otimes F(j)\cong\bigsqcup_{j\in\ob(\J)}\J(j,-)\otimes F(j)\cong\bigsqcup_{j\in\J}\bigsqcup_{\alpha\in\J(j,-)}F(j)$$The first isomorphism comes from the known description of Kan extensions as (co)ends; the middle isomorphism is precisely due to the fact $\ob(\J)$ is a disrete category with no nontrivial arrows to think about, so coends are just coproducts, and the final isomorphism is by definition of the tensor. The tensor is a bifunctor and the coend is also a (bi)functor so this automatically describes how $\lan_\iota$ works on natural transformations and morphisms.
To get un-stuck, we should think about the (proof of the) Yoneda lemma. Evaluating at $1:j\to j$ (in the coproduct $\bigsqcup_{j\in\ob(\J)}\bigsqcup_{\alpha\in\J(j,-)}$) and using naturality to extend will do the trick. Before I do so, I should mention your notation of $\bigsqcup i^x_{ab}$ is reasonable but maybe not appropriate: since coproduct is a functor, $\bigsqcup i^x_{ab}$ should be a map $\bigsqcup(\text{stuff})\to\bigsqcup(\text{other stuff})$ which comes from the universal property applied to the composition of the $i$ with the inclusion arrows. However, you're using $\bigsqcup$ to denote the map arising from the universal property directly (so, for maps $f_i:X_i\to X$, getting a map $\bigsqcup_iX_i\to X$) and this is best denoted with some other notation: I like $\langle f_i\rangle_i$ personally.
Suppose $\mu:\lan_\iota F\implies G$. Fix $j\in\J$. We know from $U(\mu)\eta=\nu$ that $\mu^j_{j,1}=\nu^j_{j,1}$. Suppose $k\in\J$ and $\alpha:j\to k$ is an arrow. $\lan_\iota(F)(\alpha)$'s restriction to $F(j)$ on the $(j,1)$th component is just the $(j,\alpha\circ1)=(j,\alpha)$th inclusion, so $\mu^k_{j,\alpha}$ is the restriction of $\mu^k\lan_\iota(F)(\alpha)=G(\alpha)\mu^j$ i.e. $\mu^k_{j,\alpha}=G(\alpha)\mu^j_{j,1}=G(\alpha)\nu^j_{j,1}$, and the right hand side is just the restriction of $G(\alpha)\nu^j=\nu^k\lan_\iota(F)(\alpha)$ to the $(j,1)$th position... which is just $\nu^k_{j,\alpha}$.
Thus $\mu=\nu$ as $j,k,\alpha$ are arbitrary and maps out of coproducts are totally determined by their components.