Consider the Dirichlet generating series of $\tau^2(n)$
$$L(s) = \sum_{n\ge 1}\frac{\tau^2(n)}{n^s}.$$
Since $\tau^2(n)$ is multiplicative it has Euler product
$$L(s) = \prod_p
\left(1+ \frac{2^2}{p^s} + \frac{3^2}{p^{2s}} + \frac{4^2}{p^{3s}}+\cdots\right).$$
Now observe that
$$\sum_{k\ge 0} (k+1) z^k = \frac{1+z}{(1-z)^3} =
\frac{1-z^2}{(1-z)^4}.$$
This gives for the Euler product that
$$L(s) = \prod_p \frac{1-1/p^{2s}}{(1-1/p^s)^4} = \frac{\zeta^4(s)}{\zeta(2s)}.$$
We can now predict the first terms of the asymptotic expansion of
$$q_n = \sum_{k=1}^n \tau^2(k)$$ using the Mellin-Perron summation formula, which gives
$$q_n = \frac{1}{2} \tau^2(n)
+ \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} L(s) \frac{n^s}{s} ds.$$
The contribution from the pole at $s=1$ is
$$\mathrm{Res}\left(L(s) \frac{n^s}{s}; s=1\right)
= \frac{1}{\pi^2} n \log^3 n
+ \left(\frac{12\gamma-3}{\pi^2} -\frac{36\zeta'(2)}{\pi^4}\right) n\log^2 n +\cdots
\\ \approx
0.101321183642338 \times n \log^3 n + 0.744341276391456\times n \log^2 n
\\+ 0.823265208269489\times n \log n + 0.460323372258732 \times n.$$
The contribution from the pole at $s=0$ is
$$\mathrm{Res}\left(L(s) \frac{n^s}{s}; s=0\right) = -\frac{1}{8}$$
but we will not include it here because it lies to the left of the zeros of the $\zeta(2s)$ term on the line $\Re(s) = 1/4.$
This gives the following asymptotic expansion:
$$\frac{1}{n} \sum_{k=1}^n \tau^2(k) \sim \frac{1}{2n} \tau^2(n)
+ \frac{1}{\pi^2} \log^3 n
+ \left(\frac{12\gamma-3}{\pi^2} -\frac{36\zeta'(2)}{\pi^4}\right) \log^2 n +\cdots.$$
This approximation is excellent, as the following table shows.
$$\begin{array}{l|ll}
n & q_n/n & \text{approx.} \\
\hline
100 & 30.46 & 30.3377762704858 \\
400 & 54.33 & 54.1863460568776 \\
1000 & 75.083 & 75.1903114374140 \\
5000 & 124.1196 & 124.110890836637 \\
\end{array}$$
Addendum. The reader is invited to supply a rigorous proof of the asymptotic expansion from above.
