How to determine net charge of a polycation using Wade's rules?
As far as I can tell, your calculations are correct. A closo structure would require $2n+2$ cluster electrons (or $4n+2$ total electrons), which is satisfied by the formula $[\ce{Bi}_n]^{(n-2)+}$. As such, for $n  = 6$, a closo octahedral structure would correspond to the ion $\ce{[Bi_6]^4+}$ as you have outlined.

The contradiction seems to be that the ion $\ce{[Bi_6]^2+}$ is not closo, but rather nido, according to e.g. Hampel &amp; Ruck:[1]


  The $\ce{[Bi_6]^2+}$ cation, a nido cluster with 16 skeletal electrons, has the shape of a distorted octahedron with an opened edge.


These require $2n$ cluster electrons for complete population of the bonding molecular orbitals, consistent with the formula $\ce{[Bi_6]^2+}$. I do not currently have access to Holleman-Wiberg (nor would I understand the German version!), so I cannot check what they have written, but it is possible that it is a simple typo.

Interestingly, some of the higher bismuth polycations also display a tendency to have "extra" electrons. For example, Greenwood and Earnshaw[2] describe the following cations:

$$\begin{array}{cccc}\hline
\text{Cation} &amp; \text{Formal oxidation state} &amp; \text{Cluster structure} &amp; \text{Point group symmetry} \\ \hline
\ce{Bi+} &amp; 1.00 &amp; - &amp; - \\
\ce{Bi3+} &amp; 0.33 &amp; \text{Triangle} &amp; D_\mathrm{3h} \\
\ce{Bi5^3+} &amp; 0.60 &amp; \text{Trigonal bipyramid} &amp; D_\mathrm{3h} \\
\ce{Bi8^2+} &amp; 0.25 &amp; \text{Square antiprism} &amp; D_\mathrm{4h} \\
\ce{Bi9^5+} &amp; 0.56 &amp; \text{Tricapped trigonal prism} &amp; C_\mathrm{3h} (\sim\!D_\mathrm{3h}) \\ \hline
\end{array}$$

The rationalisation which they provide for this is


  [The application of Wade's rules] would account for the stoichiometries $\ce{Bi3+}$ and $\ce{Bi5^3+}$ but would also lead one to expect $\ce{Bi8^6+}$ and $\ce{Bi9^7+}$ for the larger clusters. However, these charges are very large and it seems likely that the lowest-lying nonbonding orbital would also be occupied in $\ce{(Bi+)_n^2-}$. For $\ce{(Bi+)8^2-}$ this is an $\mathrm e_1$ orbital which can accommodate 4 electrons, thereby reducing the charge from $\ce{Bi8^6+}$ to $\ce{Bi8^2+}$ as observed. In $\ce{(Bi+)9^2-}$ the lowest nonbonding orbital is $\mathrm a_2''$ which can accommodate 2 electrons, thus reducing the charge from $\ce{Bi9^7+}$ to $\ce{Bi9^5+}$ as observed.


References:


Hampel, S.; Ruck, M. Synthesen, Eigenschaften und Kristallstrukturen der Cluster-Salze Bi6[PtBi6Cl12] und Bi2/3[PtBi6Cl12]. Z. Anorg. Allg. Chem. 2006, 632 (7), 1150–1156. DOI: 10.1002/zaac.200500379.
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth–Heinemann: Oxford, U.K., 1997, pp 590–591.