Let's see what happens when we try to define the model unambiguously. We can do this in several different ways. I'll consider one way.
Defining the model
I'll treat $\phi(x)$ as being real-valued field (non-compact target space). I'll work in 2d euclidean space with finite volume, saving the infinite-volume limit for later. I'll use periodic boundary conditions in both of the two dimensions, with period $L$. Thanks to the periodic boundary conditions, the action shown in the question might as well be $S\equiv\int d^2x\ (\partial\phi)^2$, where $\partial\phi$ is the gradient of $\phi$.
The correlation function should be something like
$$
\newcommand{\pl}{\partial}
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}
\la\phi(x)\phi(y)\ra
\sim \int [d\phi]\ e^{-S[\phi]}\phi(x)\phi(y),
\tag{1}
$$
but this is ill-defined because $S$ is invariant under the change of variable $\phi(x)\to\phi(x)+c$ for any constant $c$. In other words, $S$ is independent of the zero mode of $\phi(x)$. The naïve functional integral includes an integral over the zero mode, and since the integrand is not a bounded function of the zero mode, the integral does not converge.
Thanks to the finite-volume condition, this problem has an easy solution. In finite volume, the set of possible wavenumbers is discrete, so we can write
$$
\phi(x)=\frac{1}{L^2}\sum_p e^{ip\cdot x}\tilde\phi(p).
\tag{2}
$$
The zero mode (the mode with wavenumber $p=(0,0)$) is
$$
\tilde\phi(0)=\int d^2x\ \phi(x).
\tag{3}
$$
We can define (1) just by omitting the zero mode in (2), so that the original $\phi(x)$ is replaced with
$$
\phi(x)=\frac{1}{L^2}\sum_{p\neq (0,0)} e^{ip\cdot x}\tilde\phi(p).
\tag{4}
$$
Now we can define (1) by
$$
\la\phi(x)\phi(y)\ra
\propto \int \left(\prod_{p\neq (0,0)}d\tilde\phi(p)\right)\
e^{-S[\phi]}\phi(x)\phi(y),
\tag{5}
$$
with $\phi$ given by (4). Omitting the zero mode doesn't affect the action $S$, which was already independent of the zero mode. However, omitting the zero mode does affect the rest of the integrand, and that's why we did it: now the integral converges.
A side-effect of using (4) is that $\phi(x)$ is no longer a local operator, because (4) and (2) differ by the nonlocal quantity (3), and "local" is defined by the original version (2). Keep this in mind when considering the infinite-volume limit, below.
The infinite-volume limit
To understand how the definition (5) leads to the expression $\ln|x-y|^2/\mu^2$ shown in the question, consider the infinite-volume limit. Before taking the infinite-volume limit, the correlation function (5) is a periodic function of both $x$ and $y$, with period $L$ in both dimensions. Explicitly,
$$
\la\phi(x)\phi(y)\ra
\propto \frac{1}{L^2}\sum_{p\neq (0,0)}\frac{e^{ip\cdot(x-y)}}{p^2},
\tag{6}
$$
where the components of $p$ are integers times $2\pi/L$. We can write this as
$$
\la\phi(x)\phi(y)\ra
\propto
\lim_{m\to 0}
\left(\frac{1}{L^2}\sum_{p}\frac{e^{ip\cdot(x-y)}}{p^2+m^2}
-\frac{1}{L^2m^2}\right)
\tag{7}
$$
where now the sum is over all wavenumbers $p$ because the zero-wavenumber term is explicitly subtracted. In the limit $L\to\infty$, the quantity in large parentheses becomes
$$
\left(\frac{1}{L^2}\sum_{p}\frac{e^{ip\cdot(x-y)}}{p^2+m^2}
-\frac{1}{L^2m^2}\right)
\to
\int\frac{d^2p}{(2\pi)^2}\ \frac{e^{ip\cdot(x-y)}}{p^2+m^2}
\propto
\int\frac{d^2p}{(2\pi)^2}\ \frac{e^{ip\cdot(x-y)m}}{p^2+1}.
\tag{8}
$$
Notice that the limit $m\to 0$ is no longer defined: the limits $L\to\infty$ and $m\to 0$ don't commute. At this point, we have two options:
We can change the rules in the middle of the game and decide that we don't really need to take the limit $m\to 0$ after all. For $|x-y|m\ll 1$, this leads to the result shown in the question with $\mu\propto 1/m$. (Also see this post).
We can accept that $\la\phi(x)\phi(y)\ra$ doesn't have an infinite-volume limit. That shouldn't bother us, because $\phi(x)$ isn't a local operator anyway. Correlation functions of local operators are the things we really care about.
The gradient $\pl\phi(x)$ is a local operator: like the action, it was already independent of the zero mode. In the infinite-volume limit, the correlation function $\la\pl\phi(x)\pl\phi(y)\ra$ has a monotonically decreasing magnitude as a function of $|x-y|$, as expected for local operators, and it's independent of the ad-hoc regulator $\mu\propto 1/m$.
The question is why $\la\phi(x)\phi(y)\ra$ doesn't have that property. At least in the approach I used here, the answer is that $\la\phi(x)\phi(y)\ra$ is undefined in the infinite-volume limit. We define it by arbitrarily keeping $m=1/\mu$ finite, and then the result shown in the question is valid for $|x-y|/\mu\ll 1$. For such values of $|x-y|$, the result shown in the question has a monotonically decreasing magnitude as a function of distance, as expected.
