Your $\tau(n)$ and $\xi(n)$ are essentially the same as the [divisor summatory function][1], often denoted by $\sigma(n)$. Indeed, we have
$$\sigma(n)=\sum_{m=1}^n\sum_{k\mid m}1=\sum_{k=1}^n\left[\frac{n}{k}\right]=2\sum_{k=1}^{\left[\sqrt{n}\right]}\left[\frac{n}{k}\right]-\left[\sqrt{n}\right]^2.$$
That is,
$$\sigma(n)=\tau(n)+n+1=2\xi(n)+2n-\left[\sqrt{n}\right]^2.$$
Note that we have rather precise estimates for $\sigma(n)$, this is what the Dirichlet divisor problem is about. See the above Wikipedia page for more details. For example, Huxley (2003) proved for any $\varepsilon>0$ that
$$\sigma(n)=n\log n+(2\gamma-1)n+O_\varepsilon(n^{131/416+\varepsilon}).$$
In particular, these functions $\sigma$, $\tau$, $\xi$ are **not sensitive** to their arguments being prime or not prime.

**Added.** In my response, $[x]$ denotes the integral part of $x$, not the fractional part as in the OP's post. Sorry about that. At any rate, it is straightforward to relate the sum of the integral parts of $n/k$ and the sum of the fractional parts of $n/k$, because the sum of $n/k$ has a well-known asymptotic expansion.

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