This is a generalization of this question.
An $(n,k)$ partial permutation is an injection from $[k]$ to $[n]$. It can be thought of as word of length $k$ in symbols in $[n]$ without duplications.
An $(n,k)$ partial derangement is an $(n,k)$ partial permutation where the $i$th symbol is not $i$.
A good $(n,k)$ partial permutation is a $(n,k)$ partial permutation that when $n+1$ is appended to the $k+1$st position the resulting word never has $i+1$ directly following $i$. In other words if the we represent the extended partial permutation as a function $\sigma$ then for any $j$, $\sigma(j)+1 \ne \sigma(j+1)$.
It is not hard to show by inclusion-exclusion that there are the same number of good partial permutations as there are partial derangements. Is there a simple bijection?
Also, if the number of either is $d_{n,k}$, then $d_{n,k}$ satisfies two recurrences. $$d_{n,k} = (n-1) d_{n-1,k-1} + (k-1)d_{n-2,k-2}$$ and $$(n-k) d_{n,k} = d_{n,k+1} + d_{n-1,k}$$
Do either of these have nice combinatorial descriptions for good partial permutations?
To clarify: If we let $\text{pd}(n,k)$ be the partial derangements, and $\text{gpp}(n,k)$ the good partial permutations, then $$ \text{pd}(4,0)=\text{gpp}(4,0)=\emptyset$$
$$\text{pd}(4,1)=\{2,3,4\}$$ $$\text{gpp}(4,1)=\{1,2,3\}$$
$$\text{pd}(4,2)=\{21,23,24,31,34,41,43\}$$ $$\text{gpp}(4,2)=\{13,21,31,32,41,42,43\}$$
$$\text{pd}(4,3)=\{214,231,234,241,312,314,341,342,412,431,432\}$$ $$\text{gpp}(4,3)=\{132,142,143,213,241,243,321,413,421,431,432\}$$
Note that $312\notin \text{gpp}(4,3)$ as it contains 12 consecutively and $314\notin \text{gpp}(4,3)$ as if we append 5 to get 3145 we have 4 and 5 consecutively.
Edit: Added the combinatorial explanations of the recurrences for partial derangements.
To see that the first recurrence $$d_{n,k} = (n-1) d_{n-1,k-1} + (k-1)d_{n-2,k-2}$$ holds for partial derangements we describe a bijection that takes an $(n,k)$ partial derangement $\sigma$ to a pair $(\hat\sigma,l)$ where $\hat\sigma$ is either an $(n-1,k-1)$ partial derangement and and $l$ is an integer in $[n]\setminus\{k\}$ or $\hat\sigma$ is an $(n-2,k-2)$ partial derangement and $l$ is an integer an integer in $[k-1]$.
If the partial derangement does not contain $n$ then $\hat{\sigma}$ is $\sigma$ with the last element removed and $l$ is the last element. If the partial permutation contains $n$ but not in the position indexed by the last element, then $\hat{\sigma}$ is $\sigma$ with the last element removed and $n$ replaced with the last element, and $l$ is the last element. Otherwise, $l$ is the last element, and $\sigma(l)=n$, and we get $\hat{\sigma}$ by removing both $n$ and the last element and decrementing all values greater than $l$ by 1.
For example, if $n=5$ and $k=3$ $$4312\Leftrightarrow (431,2)$$ $$4351\Leftrightarrow (431,1)$$ $$4512\Leftrightarrow (31,2)$$
It is not hard to see that all of these operations are reversible, and so the given operation is a bijection.
To see that the second recurrence $$(n-k) d_{n,k} = d_{n,k+1} + d_{n-1,k}$$ holds for partial derangements, we describe a bijection that takes a pair consisting of an $(n,k)$ partial derangement $\sigma$ and an integer $l$ not occurring in $\sigma$ (there are $n-k$ possibilities for $l$) and constructs either an $(n,k+1)$ partial derangement, or an $(n-1,k)$ partial derangement $\hat\sigma$.
If $l \ne k+1$ then $\hat\sigma$ is obtained by appending $l$ to $\sigma$ giving a valid partial permutation. Otherwise, let $\hat\sigma$ be the partial derangement obtained by subtracting 1 from all the elements of $\sigma$ greater than $k+1$. Again, it is not hard to see that this gives a bijection. This seems like it should be an easy argument to transform to the same recurrence for good partial permutations, but I'm not seeing it.