I don't see how this can be obtained directly from Vandermonde's identity, but it is an easy induction over $t$ using ${k \choose j}={k-1 \choose j-1}+{k-1 \choose j}$. Grouping neighboring terms together, you will see that the LHS is just the sum of iterated differences of neighboring binomial coefficients, and iterating them $t$ times, it all boils down to ${k-t \choose m-t}$. Rather than writing it formally (who will ever read it?), I suggest you try it out with values like $k=10,m=8,t=3$.

If we put $n:=m-t$, we get as a by-product for $r=1,...,n$ $$\sum_{j=0}^{n} (-1)^{j}{m-r \choose j}{m-j-1 \choose m-n-1}=0.$$