Stirling numbers of second kind, but no two adjacent numbers in same part.

I will denote ${[n]\brace k}=\{\pi \vdash [n]:|\pi|=k\}$ the partitions of $[n]$ into $k$ blocks and i will denote $\mathbb{H}(n,k)=\{\pi \in {[n]\brace k}: \pi \text{ has no adjacent elements}\}$ so that $|\mathbb{H}(n,k)|=h(n,k).$
Consider the following function $$\varphi :{[n-1]\brace k-1}\longrightarrow \mathbb{H}(n,k),$$ given by $\varphi (\pi)=\gamma$ where if $\pi = \{B_1,\cdots ,B_k\}$ then $\gamma$ is taking each block $B$ of $\pi$ and applying the algorithm find biggest $i\in B$ such that $i,i-1 \in B$ take $B\setminus \{i-1\}$ and add $i$ to a new block that contains $n.$ in other words you send the elements that contradict your assumption of being adjacent to a block that contains $n.$
Example: $$\varphi ({\color{red}{1}24|3})=\color{red}{15}|24|3$$ $$\varphi ({\color{red}{1}2|\color{red}{3}4})=\color{red}{135}|2|4$$ $$\varphi ({1\color{red}{3}4|2})=14|2|\color{red}{35}$$ $$\varphi({1\color{red}{2}3|4})=13|\color{red}{25}|4$$ $$\varphi({1|2\color{red}{3}4})=1|24|\color{red}{35}$$ Show that this and yours are inverse of each other.


Edit: I see you want another way. Think the following. $$\mathbb{H}(n,k)={[n]\brace k}\setminus \bigcup _{i=1}^{n-1}A_i,$$ where $A_i = \{\pi \in {[n]\brace k}:i,i+1\text{ share block}\}$ So using inclusion-exclusion principle, you end up with $$h(n,k)=\sum _{i = 0}^{n-1}(-1)^i\binom{n-1}{i}{n-i\brace k}.$$ This last thing because $|A_i|={n-1\brace k}$ by collapsing $i$ and $i+1$ to one element. Then $|A_i\cap A_j|={n-2\brace k}$ and so on.

Independently show that $${n\brace k}=\sum _{i = 0}^{n-1}\binom{n-1}{i}{n-1-i\brace k-1},$$ by choosing the elements that go with $n$ in its block. Notice that this is a binomial transformation and so you can invert it as $${n-1 \brace k-1}=\sum _{i = 0}^{n-1}(-1)^i\binom{n-1}{i}{n-i\brace k}.$$ And so $h(n,k)={n-1\brace k-1}.$

I will illustrate Phicar's bijection in more detail and explain why it is invertible.

You start with a partition of $[n-1]$ into $m-1$ non-distinct parts. Let us focus on a single part. For example, when $n=12$, one part could be $$ \{1,2,3,5,6,8,9,10,11\} $$ Now, break this into chains of consecutive integers. $$ \{ 1,2,3\quad 5,6\quad 8,9,10,11 \} $$ Within each chain, we will keep the highest element, remove the second highest, keeps the third highest, remove the fourth highest, etc. The removed elements will all be put into a new part with the added element, $n$. $$ \{ 1,\color{red}2,3\quad \color{red}5,6\quad \color{red}8,9,\color{red}{10},11 \}\\\Downarrow\\\{1,3\quad6\quad 9,11\}\quad,\quad \{2,5,8,10,12\}$$ We do this for every part. It is easy to see the result will have no consecutive integers in the same part.

Now, why is this invertible? Given a partition of $[n]$ into $m$ distinct parts with no two adjacent elements in the same part, look at the part containing $n$. Everything in that part was moved there from a different part. But it is easy to see where it was moved from; the number $k$ must have come from the part containing $k+1$. After moving all these elements back, and deleting $n$, we get a partition of $[n-1]$ into $[m-1]$ parts.