The conjecture is true. For $t\in[0,1]$, let $N(t)$ be the number of 
$k\in\{1,2,\dotsc,n\}$ satisfying $\{n/k\}<t$. On the one hand,
$$\sum_{k=1}^n f\left(\frac{n}{k}\right)=\int_0^{1/2}\bigl(N(t+1/2)-N(t)\bigr)\,dt.$$
On the other hand, for any positive integer $M$,
$$N(t)=\sum_{m=1}^\infty\left(\biggl\lfloor\frac{n}{m}\biggr\rfloor-\biggl\lfloor\frac{n}{m+t}\biggr\rfloor\right)=\sum_{m=1}^M\left(\frac{n}{m}-\frac{n}{m+t}\right)+O\left(M+\frac{n}{M}\right).$$
It follows that
\begin{align*}
\sum_{k=1}^n f\left(\frac{n}{k}\right)
&=\sum_{m=1}^M\int_0^{1/2}\left(\frac{n}{m+t}-\frac{n}{m+t+1/2}\right)\,dt+O\left(M+\frac{n}{M}\right)\\
&=\sum_{m=1}^M n\log\left(\frac{(2m+1)^2}{(2m)(2m+2)}\right)+O\left(M+\frac{n}{M}\right)\\
&=n\log\left(\prod_{m=1}^M\frac{(2m+1)^2}{(2m)(2m+2)}\right)+O\left(M+\frac{n}{M}\right).
\end{align*}
By [Wallis's product][1],
$$\prod_{m=1}^\infty\frac{(2m+1)^2}{(2m)(2m+2)}=\frac{4}{\pi},$$
hence also
$$\prod_{m=1}^M\frac{(2m+1)^2}{(2m)(2m+2)}=\frac{4}{\pi}\prod_{m=M+1}^\infty\frac{(2m)(2m+2)}{(2m+1)^2}=\frac{4}{\pi}\left(1+O\left(\frac{1}{M}\right)\right).$$
Taking the logarithm of both sides, and going back to the $k$-sum,
$$\sum_{k=1}^n f\left(\frac{n}{k}\right)=n\log\left(\frac{4}{\pi}\right)+O\left(M+\frac{n}{M}\right).$$
Finally, we choose $M=\lfloor\sqrt{n}\rfloor$ to conclude that
$$\sum_{k=1}^n f\left(\frac{n}{k}\right)=n\log\left(\frac{4}{\pi}\right)+O\left(\sqrt{n}\right).$$

  [1]: https://en.wikipedia.org/wiki/Wallis_product