The following equations can be easily derived:
$$N - (q^k + n^2) + 1 = \sigma(q^{k-1})(q-1)(n+1)(n-1)$$
$$\sigma(n^2) = {q^k}\cdot\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)$$
$$n^2 = {\frac{\sigma(q^k)}{2}}\cdot\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)$$
From the last two equations, it can be proved that
$$\gcd(n^2, \sigma(n^2)) = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})},$$
and
$$\sigma(n^2) - n^2 = \left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)\cdot\left(q^k - \frac{\sigma(q^k)}{2}\right)$$
$$= \left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)\cdot\left(\frac{q^k - \sigma(q^{k-1})}{2}\right)$$
$$= \gcd(n^2, \sigma(n^2))\cdot\left(\frac{q^k - \sigma(q^{k-1})}{2}\right).$$
This last equation expresses a relationship among the quantities
$$\sigma(n^2) - n^2,$$
$$\gcd(n^2, \sigma(n^2)),$$
$$\sigma(q^{k-1}),$$
and
$$q^k.$$
Also, note that we actually have
$$\sigma(n^2) - n^2 = \frac{\left(2n^2 - \sigma(n^2)\right)\left(2q^k - \sigma(q^k)\right)}{2\sigma(q^{k-1})},$$
and
$$\sigma(n^2) - n^2 = \left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)\cdot\left(q^k - \frac{\sigma(q^k)}{2}\right)$$
$$= \left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)\cdot\left(\frac{q^k - \sigma(q^{k-1})}{2}\right)$$
$$= \gcd(n^2, \sigma(n^2))\cdot\left(\frac{q^k - \sigma(q^{k-1})}{2}\right).$$
This last equation expresses a relationship among the quantities
$$\sigma(n^2) - n^2,$$
$$\gcd(n^2, \sigma(n^2)),$$
$$\sigma(q^{k-1}),$$
and
$$q^k.$$
Also, note that we actually have
$$\sigma(n^2) - n^2 = \frac{\left(2n^2 - \sigma(n^2)\right)\left(2q^k - \sigma(q^k)\right)}{2\sigma(q^{k-1})},$$
so that
$$A(n^2) + \sigma(n^2) = \frac{{D(n^2)}\cdot{D(q^k)}}{\sigma(q^{k-1})},$$ where $A(x) = \sigma(x) - 2x$ is the abundance of $x$, and $D(x) = 2x - \sigma(x)$ is the deficiency of $x$.
In particular, we know that
$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$
and
$$\frac{\sigma(q^k)}{n^2} = 2\cdot\left(\frac{\sigma(q^{k-1})}{2n^2 - \sigma(n^2)}\right).$$
Notice that, since $\sigma(q^k)\sigma(n^2) = \sigma(N) = 2N = 2{q^k}{n^2}$ and $\gcd(q^k,\sigma(q^k)) = 1$, then$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$
is an odd integer.
The following papers obtain (ever-increasing) lower bounds for $\sigma(n^2)/q^k$:
(1) J. A. B. Dris, The Abundancy Index of Divisors of Odd Perfect Numbers, https://cs.uwaterloo.ca/journals/JIS/VOL15/Dris/dris8.html
(2) J. A. B. Dris and F. Luca, A note on odd perfect numbers, http://arxiv.org/pdf/1103.1437v3.pdf
(3) F. J. Chen and Y. G. Chen, On Odd Perfect Numbers, http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=8738171
(4) K. A. Broughan, D. Delbourgo, and Q. Zhou, Improving the Chen and Chen result for odd perfect numbers, http://www.emis.de/journals/INTEGERS/papers/n39/n39.pdf
(5) F. J. Chen and Y. G. Chen, On the index of an odd perfect number, http://zmath.sub.uni-goettingen.de/ZMATH/serials/en/search/zmath/?q=an:1301.11003&format=complete
From this M. Sc. thesis, we know that
$$\frac{11}{3} \leq \frac{\sigma(q^k)}{n^2} + \frac{\sigma(n^2)}{q^k}.$$
