$b(n,k)\ge2k$ 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.

$b(n,k)\ge2k$ 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.)

Edit: Here is a proof of the fact that $2k+8$ is enough. (All refs are to the above paper.) It suffices to construct an odd lattice $T$ (the prospective orthogonal complement) with inertia indices $(1,\sigma_-)$ and discriminant $k\langle\frac1n\rangle$, cf. Proposition 1.6.1 (where discriminant are considered with bilinear forms only, disregarding the quadratic form). Let $r=1+\sigma_-$ be the rank. We start with an even lattice $T$, see Theorem 1.10.1. (If $n$ is even, we can play with recovering the quadratic discriminant form; if $n$ is odd, it is unique.) Since signature $\bmod 8$ is determined by the discriminant form (Brown invariant), we may have to add up to $7$ to $\sigma_-$ to make it right. Next, since the length of the discriminant form is $k$, Nikulin's theorem will guarantee the existence (provided that the signature is right) whenever $r>k$. Thus, we can always find an even lattice $T$ of rank at most $k+7$. Now, just add $[-1]$ to make it odd (of rank at most $k+8$). By Proposition 1.6.1, there is a primitive extension of your lattice $S$ to a unimodular lattice such that $S^\perp\cong T\oplus[-1]$.

If you are fighting for every single unit, one can compute more precisely the Brown invariant (which depends on $k$ and the prime factors of $n$), there is no need in $[-1]$ if $n$ is odd, and, if $n$ is even, one can probably try to avoid that $[-1]$ by a "wrong" gluing of the two lattice (choosing an anti-isometry of the bilinear forms which is not an anti-isometry of quadratic ones). However, constructing an odd form $T$ in the first place would probably give a better result; unfortunately, I do not know any existence theorems for this case. And, as you've correctly noticed yourself, there is no way to improve that more than by $8$ :)