I have resolved my confusion. I post this answer for my later reference - to take paper notes - and for the aid of anyone else with a similar confusion. Unfortunately, I've just discovered that the final two sections of this answer (showing compatibility with two alternative definitions of skeleton and coskeleton) are incorrect. They can be handled with my answer here.
$\newcommand{\lan}{\operatorname{Lan}}\newcommand{\ran}{\operatorname{Ran}}\newcommand{\set}{\mathsf{Set}}\newcommand{\op}{\mathsf{op}}\newcommand{\sk}{\operatorname{sk}}\newcommand{\tr}{\operatorname{tr}}\newcommand{\cosk}{\operatorname{cosk}}\newcommand{\nat}{\mathsf{Nat}}$I will use the following slightly cleaner diagram for elaboration.
It really is just 'formal', as Zhen commented. First of all, $S:\Delta_{\le n}^\op\to\set$ is some functor (what do we call these, $n$-simplicial sets?) and $S'=\lan_n(S)$, a true simplicial set / object. The coequaliser definitely exists, so we don't actually need to worry about calculating it in general!
When $m\le n$, we can say $S'_m\cong S_m$:
Let $\pi$ be defined by $\pi\circ\iota_{g:m\to k}:=S_g:S_k\to S_m$. If $g=fh$ for $g,f,h$ all arrows in $\Delta_{\le n}$, then $\pi\circ\iota_g=\pi\circ\iota_h\circ S_f$ holds by the contravariant functoriality of $S$, so $\pi$ agrees with the source/target arrows. $\pi$ is a coequaliser:
If $q:\bigsqcup_{t:m\to k}S_k\to X$ is any other arrow that agrees with source/target, then $q=\alpha\pi$ for some $\alpha:S_m\to X$ would imply that $q\circ\iota_{1:m\to m}=\alpha$, so $\alpha$ exists uniquely if it exists at all. But if we define $\alpha:=q\circ\iota_{1:m\to m}$, then for any $g:m\to r$ ($r\le n$): $$\alpha\pi\circ\iota_g=q\circ\iota_{1:m\to m}\circ S_g=q\circ\iota_{g\circ 1:m\to r}=q\circ\iota_g$$As $q$ agrees with source/target. By the universal property of the coproduct, it must be that $q=\alpha\pi$. Therefore, $(\pi,S_m)$ is a valid coequaliser.
I was worried about explicitly calculating the coequaliser, but I don't need to. The coequaliser just supplies us with some objects we can freely attach to $S$ to have a full simplicial object $S'$, in a 'nice' way. We can begin the formal extension process by defining $S'_m:=S_m$ when $m\le n$ (valid, as shown) and $S'_f:=S_f$ whenever $f$ is an arrow in $\Delta_{\le n}^\op$. We can also define, for $k\le n$ and any $m$, $t\in\Delta(m,k)$ whatsoever, the arrow $S'_t:=\pi\circ\iota_t:S_k=S'_k\to S'_m$. Then the condition on $\pi$ reads: $$S'_g=S'_h\circ S_f=S'_h\circ S'_f$$Whenever $g=fh$ as in the diagram. It is clear that $S'$ acts functorially (contravariant) among such arrows. However we need this action to be functorial among all possible arrows in $\Delta_{\le n}$.
The Kan extension process produces, for a generic $x:m\to m'$ in $\Delta$, an arrow $\lan_n(S)(x):S'_{m'}\to S'_m$ as uniquely specified by the condition $\lan_n(S)(x)\circ\pi\circ\iota_h=\pi\circ\iota_{hx}$ for all $h:m'\to k$ with $k\le n$. That reads: $$\lan_n(S)(x)\circ S'_h=S'_{hx}$$So we see that, if $m'\le n$, $\lan_n(S)(x)=S'_x$ is forced by uniqueness. So there is no ambiguity in declaring $S'_x:=\lan_n(S)(x)$ for generic arrows $x$, and now we get full functoriality (by uniqueness and by the condition on $\lan_n(S)(\cdot)$).
So $S'$ is a full simplicial object (we only need a cocomplete category) and it agrees with $S$ on indices $m\le n$. The whole 'degenerate' business was opaque to me. But, with this definition of $S'$ on arrows, by very definition is (in the case of $\set$) every "simplex" of $S'_m$, for $m>n$, in the image of $S'_t$ on $k$-simplices, for $k\le n$ and $t:m\to k$ (coequalisers surject in $\set$). It can be shown that this is a sufficient condition (regardless of $t$) for the $(m>n)$-simplices to all be degenerate. The degeneracy operators are those supplied by the coequaliser.
When $S$ is the truncation of a full simplicial set $S$, $S'$ shall be $\sk_n(S)$:
This was already shown on $m\le n$. If $m>n$, then define $\pi$ uniquely by $\pi\circ\iota_t:=S_t:S_k\to S_m$. Since $S$ is functorial, the $\pi$ thus defined shall agree with source/target. Let $q:\bigsqcup_{t:m\to k}S_k\to X$ be any other arrow agreeing with source / target. The elements of $\sk_n(S_m)$ are precisely all $\sigma\in S_m$ that are of the form $S_t(\tau)$ for some $\tau\in S_k,\,t:m\to k$ and $k\le n$. If $q=\gamma\pi$ for some function $\gamma$, it must be that: $$q\circ\iota_t(\tau)=\gamma\circ S_t(\tau)=\gamma(\sigma)$$So the value of $\gamma$ on every element is forced.
But in fact, we may take $k=n$ and $t$ some surjection $m\to n$. For if such $t$ exists, we may find a surjection $t':m\to n$ and $t'':n\to k$ with $t''\circ t'=t$. Then, $S_{t'}\circ S_{t''}=S_t$, so there must be some $\tau'\in S_n$ that $S_{t'}(\tau')=\sigma$. Now fix some choice of surjective $\alpha:m\to n$ and $\tau\in S_n$, and define a function $\gamma:\sk_n(S_m)\to X$ by $\gamma(\sigma):=q\circ\iota_{\alpha}(\tau)$. We need to check this is well-defined.
Should $t:m\to k$, $\omega\in S_k$ be any pair with $k\le n$ and $S_t(\omega)=\sigma$, then, as before, we know $S_\alpha\circ S_\beta=S_t$ for some $\beta:n\to k$, so: $$\sigma=S_t(\omega)=S_\alpha(S_\beta(\omega))=S_\alpha(\tau)$$But $S_\alpha$ is injective because $\alpha$ is (split) surjective; thus, $\tau=S_\beta(\omega)$, and: $$g\circ\iota_t(\omega)=g\circ\iota_{\beta\alpha}(\omega)=g\circ\iota_{\alpha}(S_\beta(\omega))=g\circ\iota_{\alpha}(\tau)=\gamma(\sigma)$$Proving well-definedness and agreement with $q$.
Now for the co-skeleton. Take some $X:\Delta_{\le n}^\op\to\set$ and call its right Kan extension along $\Delta_{\le n}^\op\hookrightarrow\Delta_n^\op$ the simplicial set $X'$. The limit formula gives us: $$X'_m\cong\left\{(x_t)\in\prod_{t:s\to m,\,s\le n}X_s:X(f)x_{g:k\to m}=x_{gf:r\to m},\,\forall f\in\Delta_{\le n}(r,k)\right\}$$
This is identical to: $$X'_m\cong\nat(\tr_n\Delta^m,\tr_n X')$$Via the identification $x_t\sim\varphi(t)$. By the Yoneda lemma, $X'_m=X_m$ for indices $m\le n$. nLab takes the definition of "$n$-coskeletal" to be when that isomorphism holds, essentially. nLab also claims that, if $X$ is the truncation of a full simplicial set $X$, then: $$\cosk_n(X_m)=\nat(\sk_n\Delta^m,X)$$The standard $m$-simplicial set $\Delta^m$ is $m$-dimensional, so this formula holds trivially when $m\le n$ (by the Yoneda lemma). Why should it hold for $m>n$?
$\sk_n(\Delta^m)$ is the simplicial set with $\nu$-simplices the set of all $f:\nu\to m$ that factor through an arrow $f':k\to m$ where $k\le n$. We want to identify natural transformations $\varphi:\sk_n(\Delta^m)\implies X$ with natural transformations $\psi:\tr_n(\Delta^m)\implies\tr_n X$. Now, $\tr_n\sk_n(\Delta^m)=\tr_n\Delta^m$, so $\varphi$ can be mapped to $\psi=\tr_n\varphi$.
Let's take any $\psi:\tr_n\Delta^m\implies\tr_n X$. We need to show there is a unique way to extend $\psi$ to components $\psi_{\nu}:\sk_n(\Delta^m_\nu)\to X_\nu$ ($\nu>n$) that assemble into a natural transformation. Given any $f:\nu\to m$ in the $n$-skeleton, break down $f=gh$ where $g:k\to m$ has $k\le n$. If $\psi$ lifts to a natural transformation, then: $$\psi_\nu(f)=\psi_\nu\circ h_\ast(g)=X(h)\circ\psi_k(g)$$But the component $\psi_k$ is already given, so this extension is certainly unique. We just need to know that $\psi_\nu$ can be defined in this way unambiguously, that is, that $X(h)\circ\psi_k(g)=X(h')\circ\psi_{k'}(g')$ for any pairs $k,k'\le n$ with $gh=f=g'h'$.
We can write $f=g'h'$ in the canonical way, where $h':\nu\to k'$ is a surjection and $g':k'\to m$ is an injection. Given another: $$f=\nu\overset{h}{\longrightarrow}k\overset{g}{\longrightarrow}m$$With $k\le n$, it is true that $g=g'\circ\ell$ for some $\ell:k\to k'$. Because $\psi$ is natural: $$X(h)\circ\psi_k(g)=X(h)\circ\psi_k(g'\ell)=X(h)X(\ell)\circ\psi_{k'}(g')$$
But $f=gh=g'\ell h=g'h'$, so by injection it is true that $\ell h=h'$. So we can finish off: $$X(h)\circ\psi_k(g)=X(h)X(\ell)\circ\psi_{k'}(g)=X(h')\circ\psi_{k'}(g')$$
So the definition of $\psi$ on composites $f=gh$ as $X(h)\circ\psi_k(g)$ is well defined and lifts $\psi$ to a natural transformation - uniquely - from $\sk_n(\Delta^m)\implies X$.
Best Answer
The fact that every functor $F$ like yours, with cocomplete codomain, admits a (essentially unique ) extension to $sSet$ amounts to the universal property of the free cocompletion, yes; and yes, the extension (has a right adjoint, called the $F$-nerve) and is a left Kan extension, along the Yoneda embedding $y : \Delta \to {\sf sSet}$.
There is plenty of places where this is proved, but I can't help from the usual self-promotion: Theorem 3.1.1 here.
As for right extensions, that's another story: the opposite of the category of presheaves on $\Delta^{op}$, i.e. the category $[\Delta, {\sf Set}]^{op}$, exhibits the universal property of the free completion of $\Delta$, and the contravariant Yoneda embedding $y^\sharp : \Delta^{op}\to [\Delta, {\sf Set}]$ yield a continuous extension for every $G$ with complete domain.
Usually, even assuming $\sf C$ bicomplete, it is not the case that $\text{Lan}_y F \cong \text{Ran}_{y^\sharp} F$.