For a bijective interpretation, you first need a combinatorial interpretation for the numbers.
A look at the encyclopedia of integer sequences suggests the following:
$B_{n,k}$ is the number of set partitions of $n+1$ elements where the element $n+1$ cannot be in the same part with another element $\ge k$. (Note that I have adhered to your choice of indexing in the order col/rows and in your table, $n$ starts at 0 and $k$ starts at 1.)
It is easy to check that this interpretation works by regarding the case that $n+1$ does not share a part with $k-1$ (giving $B_{n,k-1}$) and the case that $n+1$ does share part with $k-1$ (giving $B_{n,k-1}$).
For $k=1$, the element $n+1$ is forced to be a singleton, so we recover $B_n$.
(And for $k=n+1$, the condition is empty, so we recover $B_{n+1}$.)
Your recursion (1) contains some typos, for that reason I won't write details, but the interpretation is straight-forward, that among a certain set of numbers (whose size corresponds to the size of the sub-triangle you are looking at), you choose those that go in the same part as $n+1$ and those that don't.
There can be no doubt that all this is well-known.
Say that a partition of $[n]$ is good if it has no singleton blocks and bad otherwise. $B_n$, the $n$-th Bell number, is the total number of partitions of $[n]$. If $b(n)$ is the number of bad partitions of $[n]$, $F(n)=B_n-b(n)$. As usual, a little data can’t hurt. By direct enumeration of $F(n)$ and $b(n)$ and a table of the Bell numbers I find
$$\begin{array}{cccc}
n&F(n)&b(n)&B_n\\ \hline
0&0&1&1\\
1&0&1&1\\
2&1&1&2\\
3&1&4&5\\
4&4&11&15\\
5&11&41&52\\
6&41&162&203
\end{array}$$
This very strongly suggests that $F(n+1)=b(n)$ for $n\ge 1$. To see why this is true, suppose first that $\pi$ is a bad partition of $[n]$; then we can form a good partition of $[n+1]$ by gathering all of the singletons of $\pi$ into a single block and putting $n+1$ into that block. Conversely, if $\pi$ is a good partition of $[n+1]$, we can form a bad partition of $[n]$ by taking the block of $\pi$ containing $n+1$, throwing away $n+1$, and converting the rest of the block to singletons. These operations are clearly inverses of each other and thus establish a bijection between bad partitions of $[n]$ and good partitions of $[n+1]$.
This immediately gives us the recurrence $F(n+1)=B_n-F(n)$. Unwrapping the recurrence, we find that:
$$\begin{align*}
F(n+1)&=B_n-F(n)\\
&=B_n-\big(B_{n-1}-F(n-1)\big)\\
&=B_n-B_{n-1}+F(n-1)\\
&=B_n-B_{n-1}+\big(B_{n-2}-F(n-2)\big)\\
&=B_n-B_{n-1}+B_{n-2}-F(n-2)\\
&\;\vdots\\
&=\sum_{i=0}^k(-1)^iB_{n-i}+(-1)^{k+1}F(n-k)\\
&\;\vdots\\
&=\sum_{i=0}^{n-1}(-1)^iB_{n-i}\;.
\end{align*}$$
Added: For the first question, let $\pi$ be a good partition of $[n+1]$, and let $B$ be the block containing $n+1$. There is at least one element of $[n]$ in that block, so $[n]\setminus B$ is a proper subset of $[n]$. If $|[n]\setminus B|=k$, $\pi\setminus\{B\}$ can be any one of the $F(k)$ good partitions of $[n]\setminus B$. Conversely, all good partitions of $[n+1]$ can be obtained by choosing a proper subset $S$ of $[n]$, forming a good partition of $S$, and adding to it the block $\{n+1\}\cup([n]\setminus S)$. Since $[n]$ has $\binom{n}k$ subsets of cardinality $k$, $[n+1]$ has $\binom{n}kF(k)$ good partitions in which the block containing $n+1$ has $n-k$ other elements, and it follows that $$F(n+1)=\sum_{k=0}^{n-1}\binom{n}kF(k).$$
Best Answer
Element $n$ must be in some part of the partition. Of the remaining $n - 1$ elements, suppose $j$ of them are in the same part as element $n$. The number of ways in which this can occur is $\binom{n-1}j$. The number of ways of partitioning the rest of the elements is $p(n - j - 1)$. Multiply those together and sum over $j$.