We have the following claim where $n\ge j$ (the sum is zero when $n\lt
j$ and the claim holds by inspection)
$$\sum_{k=j}^n
\frac{1}{k} {n\choose k-1} {k\choose j} B _{k-j}
= \delta_{nj}.$$
This is
$$\sum_{k=j}^n
{n+1\choose k} {k\choose j} B _{k-j}
= \delta_{nj} \times (n+1).$$
Now
$${n+1\choose k} {k\choose j} =
\frac{(n+1)!}{(n+1-k)! \times j! \times (k-j)!}
= {n+1\choose j} {n+1-j\choose k-j}$$
and we find
$$\sum_{k=j}^n
{n+1-j\choose k-j} B_{k-j}
= \delta_{nj} \times (n+1) \times {n+1\choose j}^{-1}$$
or
$$\sum_{k=0}^{n-j}
{n+1-j\choose k} B_{k}
= \delta_{nj} \times (n+1) \times {n+1\choose j}^{-1}
\\ = \delta_{nj} \times (n+1) \times {n+1\choose n}^{-1}
= \delta_{nj}.$$
To prove this last form we put on the LHS
$$-B_{n+1-j} + \sum_{k=0}^{n+1-j}
{n+1-j\choose k} B_{k}
\\ = -B_{n+1-j}
+ (n+1-j)! [z^{n+1-j}] \frac{z}{\exp(z)-1} \exp(z).$$
Observe that
$$\frac{z}{\exp(z)-1} \exp(z) =
\frac{z}{\exp(z)-1} (\exp(z)-1) +
\frac{z}{\exp(z)-1}
\\ = z + \frac{z}{\exp(z)-1}$$
so that we get
$$-B_{n+1-j}
+ (n+1-j)! [z^{n+1-j}] z
+ (n+1-j)! [z^{n+1-j}] \frac{z}{\exp(z)-1}
\\ = - B_{n+1-j}
+ (n+1-j)! \delta_{nj}
+ B_{n+1-j}
= \delta_{nj},$$
which is the RHS. This concludes the argument.
Using a coefficient extractor Eg
\begin{eqnarray*}
\binom{k}{m} = [x^m]:(1+x)^k.
\end{eqnarray*}
We have
\begin{eqnarray*}
\sum_{k\ge1}\binom km\binom{n-1}{k-1} &=& [x^m]: \sum_{k=1}^{n} \binom{n-1}{k-1} (1+x)^k \\
&=& [x^m]: (1+x) (2+x)^{n-1} \\
&=& 2^{n-1-m} \binom{n-1}{m} +2^{n-m} \binom{n-1}{m-1} \\
&=& 2^{n-1-m}\left( \binom{n-1}{m} +2 \binom{n-1}{m-1} \right)\\
&=& 2^{n-1-m} \frac{(n-1)!}{m!(n-m)!} \left( n-m+2m \right)\\
&=& 2^{n-1-m} \frac{(n-1)!}{m!(n-m)!} (n+m).\\
\end{eqnarray*}
Which is the first identity (upto a factor of $n$).
Best Answer
Consider binary words of length $2n$ that contain $m$ $0$'s. Index the entries with $1_a,2_a, \cdots , n_a, 1_b,2_b, \cdots , n_b$. Let $A_{00}$ be the subset of $[n]$ whose elements $i$ have both $i_a$ and $i_b$ equal to $0$ & Let $A_{01}$ be the subset of $[n]$ whose elements $i$ have one $i_a$ and $i_b$ equal to $0$ and the other equal to $1$.
$A_{00}$ has cardinality $k=0, \cdots , \left\lfloor\frac m2\right\rfloor$ and $A_{01}$ has cardinality $m-2k$ and these can be the $a$ or $b$ entry in $2^{n-2k}$ ways. So \begin{eqnarray*} \sum_{k=0}^{\left\lfloor\frac m2\right\rfloor} \binom nk\binom{n-k}{m-2k}2^{m-2k} =\binom{2n}m. \end{eqnarray*}