$\newcommand\bmod{\mathbin\%}$Note that
$$ \left\lfloor \frac{a \bmod p + b \bmod p}{p} \right\rfloor = \left\lfloor \frac{a+b}{p} \right\rfloor - \left\lfloor \frac{a}{p} \right\rfloor - \left\lfloor \frac{b}{p} \right\rfloor$$
so your expression can be written as $F_{p_k\#}(x+p_k) - F_{p_k\#}(x) - F_{p_k\#}(p_k)$, where
$$ F_n(x) := \sum_{i\mid n} \left\lfloor \frac{x}{i} \right\rfloor \mu(i).$$
Now, the expression $F_n$ can be simplified by elementary number theoretic computations. Indeed, observe that
$$ \left\lfloor \frac{x}{i} \right\rfloor = \sum_{m \leq x: i\mid m} 1$$
and thus after interchanging summations,
$$ F_n(x) = \sum_{m \leq x} \sum_{i\mid n, i\mid m} \mu(i).$$
But by Möbius inversion, $\sum_{i\mid n, i\mid m} \mu(i)$ equals 1 when $n$, $m$ are coprime and 0 otherwise. Thus $F_n(x)$ is nothing more than the number of natural numbers $m$ less than $x$ that are coprime to $n$. In particular, this gives
$$ F_{p_k\#}(x+p_k) \geq F_{p_k\#}(x)$$
and
$$ F_{p_k\#}(p_k) = 1$$
which explains your numerically observed phenomenon.
The quantities here are somewhat reminiscent of the quantity $\pi(x+y)-\pi(x)-\pi(y)$ that occurs in the second Hardy–Littlewood conjecture (which, by the way, is widely believed to be false). Indeed, the quantity $F_{p_k\#}(x)$ (that is, the number of natural numbers up to $x$ that have no prime factor less than or equal to $p_k$) is more commonly denoted $\pi(x,p_k)$ in the analytic number theory literature.
EDIT: I also encountered some related expressions and inequalities in my paper A remark on partial sums involving the Mobius function (and with more advanced versions of these inequalities also in this paper of Granville and Soundararajan: Negative values of truncations to $L(1, \chi)$).