The following question is taken from $\textit{Arrows, Structures and Functors the categorical imperative}$ by Arbib and Manes
$\text{(1):}$ Given a relation ${\stackrel{\small{R}\stackrel{\small\small p_1}{\stackrel{\longrightarrow}{\longrightarrow}}\small{A}}{\small\small p_2}}{\stackrel{\stackrel{}{\stackrel{}{}}{}}{}},$ we can define the $\textbf{the equivalence relation generated by}$ $(R,p_1,p_2)$ to be $\bar{R}$, the smallest equivalence relation containing the relation defined by $R:(a,a')\in \bar{R}$ iff either $a=a'$ or $(a,a')$ can be linked by a chain $(a_1,a_2,\ldots, a_{n+1})$ of elements of $A$ where $a=a_1,a'=a_{n+1}$ and for each $k$ with $1\leq k\leq n,$ either $(a_k, a_{k+1}) \text{ or } (a_{k+1},a_i)$ belongs to $E_R,$ where $E_R=\{(p_1(r), p_2(r))\mid r\in R\}\subset A\times A,$ and $p_k:R\rightarrow A:(a_1,a_2)\mapsto a_k, k=1,2.$
$\text{19 Corollary:}$ The colimit of any diagram $D$ in $\textbf{Set}$ is obtained as follow. Consider the disjoint union $\coprod D_i=\{(d,i)\mid d\in D_i\}.$ Let $R$ be the subset
$$\{(d,i;d',i')\mid \exists e:i\rightarrow i' \text{with } D_e(d)=d' \}$$
of $\coprod D_i \times \coprod D_i$ and let $\bar{R}$ be the equivalence relation generated by $R$ as in $(1)$. Let $\gamma_i:D_j\rightarrow (\coprod D_i)/R, d\mapsto [d,j].$ Then $((\coprod D_i)/R,(\gamma_j))$ is the colimit of $D.$
Exercise 3.2: Let $E$ be the equivalence relation on $\mathbb{N}$ whose two equivalence classes are "even" and "odd". Note that $E$ is a poset via $(n, m)\leq (n',m')$ iff $n\leq n'$ and $m\leq m'$ such that the restricted projections $p,q: E\rightarrow \mathbb{N}$ are order-preserving. Show that $coeq(p,q)$ exists in $\textbf{Poset,}$ and is not the same as in $\textbf{Set}$.
post: post which was answered by Vladimir Sotirov
$\textbf{Use exercise 3.2:}$ to conclude that the analogous construction to $\text{19 Corollary}$ does not construct colimits of diagrams in $\text{Poset}.$
My $\textbf{question}$ is, the reason for the colimit $\textbf{Poset}$ can't be constructed, is it because the Poset in exercise 3.2 don't have an initial object or because the coproduct for the Poset in rhat exercise don't exists? But coequalizer for that Poset and also coproduct are considered as colimit. The text I am using have no theorems for criterias when limits/colimits can or can not be constructed. Thank you in advance.
The exercise is unclear - it all depends on what "analogous" means.
According to nLab, the category of posets is cocomplete. Hence, all colimits exist and all coproducts and coequalisers exist. So quite literally, the coproduct-coequaliser construction (which is "analogous" in that it is the same for every category which can compute it) computes all colimits.
I am then guessing they don't mean for the "analogy" to be in terms of coproducts and coequalisers, but rather in terms of the setwise coproduct and the equivalence relation quotient. But this doesn't quite make sense, since, although you'll compute a set, you're not immediately computing a poset. So what they really mean is this:
As for your comments, I'm sorry to say they don't really make sense. You're interchanging "poset" (just, the literal thing, a set with a certain ordering) with the category of all posets.
Well, although you can view individual posets as categories in their own right, here you're asking about the category of all posets, so what it means to have a colimit is very different. Also, even if initial and terminal objects don't exist (which, by the way, they do in the category of all posets) that doesn't mean a particular colimit won't exist, it just means: "not all colimits exist".
If you just want a general reason, not for: "why don't all colimits exist?" (because they apparently do!) but rather for: "why don't all colimits in $\mathsf{Poset}$ look like the corresponding colimits in $\mathsf{Set}$?"
Then my explanation is: the universal properties for $\mathsf{Poset}$ require the unique maps to also be monotone, which means that we can't just take the same quotients that we take in $\mathsf{Set}$: how do you order the quotient set? Well, you use some ordering generated by all the quotient maps but this is quite complicated and, importantly, may make some additional identifications (by that I mean, where in $\mathsf{Set}$ we may have two different equivalence classes $a$ and $b$, we might find that $b\ge a\le b$ is forced hence $a=b$...) so this is a more 'aggressive' process, and produces different quotient objects then are produced in $\mathsf{Set}$.
A concrete example of this is exactly as in Vlad's answer. The $\mathsf{Set}$-quotient $\Bbb N/E$ has two elements, let's call them $0,1$ to refer to the even and odd classes respectively. As $2\le 3\le 4$ in $\Bbb N$ (these are random numbers) it is actually forced that $0\le 1\le 0$ ($2\sim0$ because it is even, etc.) hence $1=0$ - so there are no longer two distinct equivalence classes, there is only one. A new "identification" has been made.