Suppose:
- $C$ is a discrete category with two objects $c_1,c_2$
- $D$ is a category with two objects $d_1,d_2$ and one non-identity morphism $f: d_1 \rightarrow d_2$
- $F: C \rightarrow Set$ is a functor that maps $F(c_1) = \{x_1\}, F(c_2) = \{x_1,x_2\}$
- $K: C \rightarrow D$ is a functor that maps $K(c_1) = d_1, K(c_2) = d_2$
Since Set is co-complete we can use the pointwise Kan extension formula (6.2.1 in Riehl's Category Theory in Context) to compute \begin{align*} & Lan_KF(d_1) = F(c_1) = \{x_1\}\\ & Lan_KF(d_2) = F(c_1) + F(c_2) = \{x_1\} + \{x_1, x_2\}\\ & Lan_KF(f) = \iota_{\{x_1\} \subseteq \{x_1\} + \{x_1, x_2\}} \end{align*} where $+$ is the disjoint union.
Now consider the subcategory $Inc$ of $Set$ in which morphisms are restricted to inclusion functions $\iota: S \rightarrow S \cup S'$ and define the functor $F': C \rightarrow Inc$ with the same pointwise definition as $F$. $Inc$ is a preorder, and since $\{x_1\}$ is the $\iota$-smallest element of $Inc$ that is $\geq F(c_1) = \{x_1\}$ and $\{x_1, x_2\}$ is the $\iota$-smallest element of $Inc$ that is $\geq F(c_2) = \{x_1, x_2\}$ we have: \begin{align*} & Lan_KF'(d_1) = F'(c_1) = \{x_1\}\\ & Lan_KF'(d_2) = F'(c_2) = \{x_1, x_2\}\\ & Lan_KF'(f) = \iota_{\{x_1\} \subseteq \{x_1, x_2\}} \end{align*}
Is this computation correct? What is the intuition for why the Kan extension of $F$ along $K$ can change when we replace $F$ with another functor with an identical pointwise definition?
Your computations are correct. It shouldn't be so surprising, though, that the value of $Lan_K F$ has changed: this is a thing defined by a universal property, so it depends on arrows in addition to objects. The definition of a Kan extension appeals to some natural transformations, which depend on how arrows work. A functor $D \to Set$ is an arrow in $Set,$ and the universal property says that a map from $F$'s values to this arrow's endpoints is the same as a map to this arrow from the arrow $Lan_K F(D).$ If you make the target category into a poset, every diagram becomes commutative, and so a map between arrows in the same as a map between their endpoints, whence the Kan extension in this case turns out to be just an extension of $F$. Initially, this was not the case, essentially because not every square commutes.
After all, the very way you computed the values of the Kan extension reveals this dependence of arrows in the domain (I assume you mean the colimit over $K \downarrow d$): the only thing that changes is that the colimit over the category $\bullet \> \bullet$ is now different. But yes, once colimits change, Kan extensions, which in good cases are some colimits, might also change.