Let $d_{sf}(n)$ be the number of square-free divisors of $n$, and let $D_{sf}(k)=\sum_{n=1}^{k} d_{sf}(n)$ denote the corresponding summation function. Mertens showed that the asymptotic expansion of $D_{sf}(k)$ is
$$\displaystyle D_{sf}(k)=\frac{1}{\zeta(2)}k \log(k)+\ (\frac{2\gamma-1}{\zeta(2)}-\frac{2\zeta'(2)}{\zeta^2(2)})k+O(k^{\frac{1}{2}} \log(k))$$
where the error term was recently reported to be, under the Riemann hypothesis, $O(k^{\frac{1}{4}+\epsilon})$.
It is also known that, for $Re(s)>1$, the following asymptotic expansion holds:
$$\sum_{n=1}^\infty \frac{d_{sf}(n)}{n^s}=\frac{\zeta^2(s)}{\zeta(2s)}$$
I would like to determine an asymptotic expansion for the weigthed summation corresponding to the case $s=1$, i.e.
$$\sum_{n=1}^x \frac{d_{sf}(n)}{n}$$
which expresses the average number of square free divisors for $n\leq x$, when $x$ tends to $\infty$. After some calculations, I suppose that this asymptotic expansion may be given by $$\frac{1}{2\zeta(2)} \log^2(x)+O(\log(x))$$ but I would like to get a confirmation of this. Based on the growth rate of the summation, I also hypothesize that a more accurate estimate could be written as $$\frac{1}{2\zeta(2)} \log^2(x)+a \log(x)+b+O(\frac{1}{\sqrt{x}})$$ but again I would like to get a formal proof.
This is a very basic question that follows trivially via summation by parts. We have that \[\sum_{n \leq x} \frac{d_{sf}(n)}{n} = \frac{1}{x} \sum_{n \leq x} d_{sf}(n) + \int_{1}^{x} \frac{1}{t^2} \sum_{n \leq t} d_{sf}(n) \, dt. \] Using the expression \[\sum_{n \leq x} d_{sf}(n) = A x \log x + B x + E(x), \] where the error term $E(x)$ satisfies $E(x) = O(\sqrt{x} \log(x + 2))$, we see that \[\sum_{n \leq x} \frac{d_{sf}(n)}{n} = A \log x + B + O\left(\frac{\log x}{\sqrt{x}}\right) + \int_{1}^{x} \left(\frac{A \log t}{t} + \frac{B}{t} + \frac{E(t)}{t^2}\right) \, dt. \] The integral is equal to \[\frac{A (\log x)^2}{2} + B \log x + \int_{1}^{\infty} \frac{E(t)}{t^2} \, dt + O\left(\int_{x}^{\infty} \frac{\log t}{t^{3/2}} \, dt\right), \] and this last term is \[O\left(\frac{\log x}{\sqrt{x}}\right). \] So if we write \[C = \frac{A}{2}, \qquad D = A + B, \qquad E = B + \int_{1}^{\infty} \frac{E(t)}{t^2} \, dt, \] then we have the asymptotic \[\sum_{n \leq x} \frac{d_{sf}(n)}{n} = C (\log x)^2 + D \log x + E + O\left(\frac{\log x}{\sqrt{x}}\right). \]
As an aside, you write that the error term was recently reduced to $O(k^{1/2 + \varepsilon})$, but this is a weaker error term than $O(k^{1/2} \log k)$.