coequalizer of simplicial sets This is the statement whose first line of proof I am confused. This is on page 10, of Goerss, Jardine's Simplicial Homotopy Theory.

(i) How does one prove the "presentation" of $\partial \Delta^n$? 
(ii) How does one obtain the first coequalizer? 
In general, I am just confused in how to prove coequalizing diagram for simplicial sets. 

Can ignore: I have typed up my attempt for (i). I would be happy to see for a better presentation (hand drawn is great too). 

 A: Use the cosimplicial identities to check that for $i<j$ the square
$\require{AMScd}$
\begin{CD}
    [n-2]@>\delta_{j-1}>> [n-1]\\
    @VV\delta_i V   @VV \delta_i V\\
    [n-1] @>\delta_j>> [n].
\end{CD}
is a pullback in the simplex category $\bf\Delta$. As Pece points out, pullbacks in $sSet$ are computed levelwise, so since the Yoneda embedding is fully faithful, we get for $i<j$ pullbacks in $sSet$
$\require{AMScd}$
\begin{CD}
    \Delta^{n-2}@>\delta_{j-1}>> \Delta^{n-1}\\
    @VV\delta_i V   @VV \delta_i V\\
    \Delta^{n-1} @>\delta_j>> \Delta^n.
\end{CD}
Now recall that the p-simplices of $\partial \Delta^n$ are those maps $[p]\rightarrow [n]$ in $\bf\Delta$ which factor $[p]\rightarrow [n-1]\xrightarrow{\delta_i}[n]$ for some $i$. Thus there is an epimorphism $\bigsqcup_{0\leq i\leq n}\Delta^{n-1}\rightarrow \partial \Delta^n$ which is the map $\delta_i$ on the $i$-indexed component, and you check by hand using the fact that the previous squares are pullbacks in $sSet$ that 
$$\bigsqcup_{0\leq i<j\leq n}\Delta^{n-2}\rightrightarrows\bigsqcup_{0\leq i\leq n}\Delta^{n-1}\rightarrow \partial\Delta^n.$$
Hopefully this fills out the details of what you have worked out in your attempt. Another way to see this is to use the fact that every epimorphism in $sSet$ is the coequaliser of its kernel pair. Use the pullback squares to construct the kernel pair of the epimorphism and you get the result.
The other coequaliser involving the simplicial horn follows in the same manner. Of course the p-simplices of $\Lambda^n_k$ are those maps $[p]\rightarrow [n]$ in $\bf\Delta$ that factor $[p]\rightarrow [n-1]\xrightarrow{\delta_i}[n]$ for some $i\neq k$. Therefore we have an epimorphism  $\bigsqcup_{0\leq i\leq n,i\neq k}\Delta^{n-1}\rightarrow \Lambda^n_k$ which is $\delta_i$ on the $i$-indexed copy. Use the previous pullback square to calculate the kernel pair of these epimorphisms and you are done.
Edit:
Recall that the kernel pair $K(f)$ of a morphism $f:x\rightarrow y$ in a category $\mathcal{C}$ (if it exists) is the pullback of the cospan $x\xrightarrow{f}y\xleftarrow{f}x$. Note that it comes equipped with a pair of projections $\pi_1,\pi_2:K(f)\rightarrow x$. If $f$ is an effective epimorphism then it is the coequaliser in the diagram
$$K(f)\stackrel{\pi_1,\pi_2}{\rightrightarrows} x\xrightarrow{f} y.$$
In $sSet$, or more generally any topos, every epimorphism is effective, and since $sSet$ is complete we can therefore ask for the kernel pair $K(\hat\delta)$ of the epimorphism $\hat\delta=\{\delta_i\}:\bigsqcup_{0\leq i\leq n}\Delta^{n-1}\rightarrow\Delta^n$ to get a coequaliser 
$$K(\hat\delta)\stackrel{\pi_1,\pi_2}{\rightrightarrows}\bigsqcup_{0\leq i\leq n}\Delta^{n-1}\xrightarrow{\hat\delta}\Delta^n.$$
But we can calculate $K(\hat\delta)$ directly, levelwise if necessary, as the pullback of
$$\bigsqcup_{0\leq i\leq n}\Delta^{n-1}\xrightarrow{\hat\delta}\Delta^n\xleftarrow{\hat\delta}\bigsqcup_{0\leq j\leq n}\Delta^{n-1}$$
and we have a canonical model for this pullback. In particular a p-simplex in $K(\hat\delta)$ is a pair $(x,y)$ of p-simplices in $\Delta^{n-1}$ which satisfy $\delta_i(x)=\delta_j(x)$ for some $i,j$. This should seem familiar. 
Now we have the pair of maps $\bigsqcup_{0\leq i<j\leq n}\Delta^{n-2}\rightarrow \bigsqcup_{0\leq i\leq n}\Delta^{n-1}$ which you describe, and since these are coequalised by $\hat\delta$ we get a comparison map $\bigsqcup_{0\leq i<j\leq n}\Delta^{n-2}\rightarrow K(\hat\delta)$. Using what we know about the interactions of the maps $\delta_i$, that is, our pullback squares, we check directly that this comparison map is one-to-one and onto.
