Consider over $\mathbb{C}$. Let $X$ be a projective scheme and let $\mathcal{O}(1)$ be an ample line bundle on $X$. Consider Gieseker (semi)stability. For a coherent pure sheaf $\mathcal{F}$ on $X$, there is a unique filtration, called the Harder-Narasimhan filtration $$0=\mathcal{F}_{\leq 0}\subsetneq\cdots\subsetneq\mathcal{F}_{\leq l}=\mathcal{F}$$ such that $\mathcal{F}_i:=\mathcal{F}_{\leq i}/\mathcal{F}_{\leq i-1}$ is semistable and $p(\mathcal{F}_1)>\cdots>p(\mathcal{F}_l)$ where $p(\mathcal{F})$ denotes the reduced Hilbert polynomial and $>$ is the total order on $\mathbb{Q}[t]$.
For a Noetherian local $\mathbb{C}$-algebra $A$, we can define a flat $S:=\mathrm{Spec}(A)$-family of semistble $\mathcal{O}_X$-modules, as a coherent $\mathcal{O}_{X\times S}$-module $\mathcal{F}$ such that $\mathcal{F}$ is flat over $S$ and at geometric points $s\in S$, the fibre $\mathcal{F}|_{X\times\{s\}}$ is semistable. It seems that for a flat $S$-family of coherent pure $\mathcal{O}_X$-modules, there is not always a filtration $\{\mathcal{F}_{\leq i}\}$ such that it restricts to the Harder-Narasimhan filtration at geometric points.
My question is why there is no existence in general. Are there calculable examples showing this?
I'm not sure that I understand your question correctly, but if you want the filtration on $X \times S$ to be a Harder-Narasimhan filtration, I think any example where $\mathcal F$ is not semistable, but $\mathcal F_s = \mathcal F_{X \times \{s\}}$ is semistable, should do.
Suppose there is a Harder-Narasimhan filtration for $\mathcal F$ which restricts to a HN-filtration on $X \times \{s\}$. If $\mathcal F|_{s}$ is semistable, it has trivial Harder-Narasimhan filtration. So $\mathcal F_{\leq i}|_{X \times \{s\}} = 0$ for $i < \ell$. But the support of a coherent sheaf is closed, so that $F_{\leq i} = 0$ on all of $X \times S$ (The support is a closed subset $Y \subset X \times S$ which maps to a closed subset $Z = \operatorname{pr}(Y) \subset S$ by projectivity of $X$. But $Z$ does not contain the closed point $s \in S$, so $Z = \emptyset$).
So the HN-filtration of $\mathcal F$ is trivial, but if $\mathcal F$ is not semistable, this is a contradiction.