$b(n,k)\ge2k$ is a pretty good bound. You definitely have $b(n,k)\le2k+\mathrm{const}$ with a constant not very large (something like at most $8$, but you can make it better depending on $kn\bmod8$). Here is the argument.
The situation with even form is somewhat simpler and everything follows from
MR0525944 (80j:10031) Nikulin, V. V. Integer symmetric bilinear forms and some of their geometric applications. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 43 (1979), no. 1, 111–177, 238. 10C05 (14G30 14J17 14J25 57M99 57R45 58C27)
(There is an English translation.) That would work if $n$ is even: you might need to add up to $7$ to get the right signature, one more might be needed for Nikulin's existence, and yet one more to make it odd: just add $[-1]$. If $n$ is odd, I guess you can pass to the maximal even sublattice first.