Let $\mathcal{D}$ be a finite category and let $F:\mathcal{D}\to\mathbf{Set}$ be a functor. Since $\mathbf{Set}$ has (co)equalizers and finite (co)products, by the Existence Theorem $F$ has a limit and (dually) a colimit, and the Theorem tells us how to construct them (see for instance this answer for an explicit construction of the colimit).
Here is another neat way to find the limit. Recall that, for $A$ a set and $\Delta_A:\mathcal{D}\to\mathbf{Set}$ the constant functor, we have a bijection $\hom(A,\lim F)=[\mathcal{D},\mathbf{Set}](\Delta_A,F)$. Setting $A=\{ 1 \}=\ast$ to be a one-element set, we therefore find $$ \begin{aligned} \lim F &= \hom(\ast,\lim F) = [\mathcal{D},\mathbf{Set}](\Delta_\ast,F)\\ &= \Big\{ \big(f_i:\ast\to F(i)\big)_i\in \prod_{i\in\mathcal{D}} \hom\big(\ast,F(i)\big) \;\Big\vert\; f_j=F(\alpha)\circ f_i \text{ for all }\alpha:i\to j \Big\}. \end{aligned} $$ Using that $\hom(\ast,F(i))=F(i)$ for each $i$, we obtain the usual description of the limit in $\mathbf{Set}$!
Question: Is there a similar trick, for example using $\hom(\text{colim} F,A)=[\mathcal{D},\mathbf{Set}](F,\Delta_A)$, to find a description of the colimit?
My approach so far has been to first reduce the question to the case $\mathcal{D}=\{1,2\}$ and to find the right set $A$ to deduce that $F(1)\sqcup F(2)=\big(F(1)\times\{1\}\big)\cup\big(F(2)\times \{2\}\big)$, but I haven't found such a set yet. I suspect that dividing out by the equivalence relation will somehow follow from the condition $f_i=f_j\circ F(\alpha)$ on a cone $(f_i:F(i)\to A)_i$.
Here I work out @ZhenLin's comment.
The ncatlab atom page suggests that we have a functor $\DeclareMathOperator{\Atom}{Atom}\DeclareMathOperator{\BPos}{BPos} \DeclareMathOperator{\Set}{Set} \Atom : \BPos \to \Set$ from posets $(S,\leq)$ with bottom $\perp \in S$ to sets defined as $\Atom(S,\leq) := \cup\{ s \in S \mid s \neq \perp, \forall t \in S : t \leq s \Rightarrow t = s \lor t = \perp\}$. Then if $P : \Set \to \BPos$ is the power set functor we have $\Atom(P(S)) = S$.
Now suppose $F : \mathcal{D} \to \Set$ has a colimit $\varinjlim F$. Then we have $$\varinjlim F = \Atom(P(\varinjlim F)) = \Atom(\Set(\varinjlim F,\{0,1\})) = \Atom([\mathcal{D},\Set](F,\Delta_{\{0,1\}})).$$
Here we used the representability equalities $P = \Set(\bullet,\{0,1\})$ and $\Set(\varinjlim F, \bullet) = [\mathcal{D},\Set](F,\Delta_{\bullet})$.
So if we can define a bottom poset structure on $[\mathcal{D},\Set](F,\Delta_{\{0,1\}})$, then we can take as definition $\varinjlim F := \Atom([\mathcal{D},\Set](F,\Delta_{\{0,1\}})).$
We set $\sigma \leq \tau$ for $\sigma,\tau \in [\mathcal{D},\Set](F,\Delta_{\{0,1\}})$ if $\tau(d)(x) = 1 \Rightarrow \sigma(d)(x) = 1$ $\forall d \in \mathcal{D}, x \in F(d)$ and we set $\perp \in [D,\Set](F,\Delta_{\{0,1\}})$ by $\perp(d)(x) = 0$ $\forall d \in \mathcal{D},x \in F(d)$.
Then we get $\Atom([\mathcal{D},\Set](F,\Delta_{\{0,1\}}))$ to be all graphs $\sigma$ where only one path component is 1 and the rest is 0.
To get a bijection with the standard definition of colimit using equivalence relations $\Atom([D,\Set](F,\Delta_{\{0,1\}})) \leftrightarrows \coprod_{d \in \mathcal{D}} F(d) /\sim$ we can map a graph $\sigma$ to an element $x \in F(d)$ such that $\sigma(d)(x) = 1$ and conversely, we map an element $x \in F(d)$ to the graph $\sigma$ that is 1 on the path component of $x$ and 0 elsewhere.
So finally we have $$ \varinjlim F = \Atom([\mathcal{D},\Set](F,\Delta_{\{0,1\}})) \quad \text{and} \quad \varprojlim F = [\mathcal{D},\Set](\Delta_{\{*\}},F)$$ and what feels neat about this construction of $\varinjlim F$ is that we can completely describe it with subobjects so we don't need disjoint unions or equivalence relations.