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.
Rewrite the recurrence: $$B_{n+1}=\sum_{k=0}^n\binom{n}kB_k=\sum_{k=0}^n\binom{n}{n-k}B_k=\sum_{k=0}^n\binom{n}kB_{n-k}\;.$$
To form a partition of $[n+1]$, first fix the part containing $n+1$; say that it has $k$ other elements. There are $\binom{n}k$ ways to choose these other elements, and there are $B_{n-k}$ ways to partition the remaining $n-k$ elements of $[n+1]$, so there are $\binom{n}kB_{n-k}$ partitions of $[n+1]$ that have $n+1$ in a part with exactly $k$ other elements. Summing over the possible values of $k$ yields the recurrence.
There isn’t a nice closed form; see Wikipedia for a not so nice expression involving Stirling numbers of the second kind and for asymptotic results.
Best Answer
For concreteness, let's suppose we are partitioning the set $\{1, 2, \dots, n+1\}$. Focus first on the block containing the element $1$. Let $k$ denote the number of elements other than $1$ that belong to this block. We can choose these elements in $\binom{n}{k}$ ways. Having formed this block, we partition the remaining $n + 1 - (k + 1) = n -k$ elements in $B_{n-k}$ ways. Summing over $k$ gives $$ \sum_{k = 0}^n \binom{n}{k} B_{n-k}. $$ By the symmetry of the binomial coefficients, this expression is equivalent to $$ \sum_{k = 0}^n \binom{n}{k} B_{k}. $$