You have a good intuition on the possible answer because it does involve oscillations, but you need to visualize the string differently. The origin of the conundrum is not quantum, but relativistic. Here's why:
Consider an inertial frame O and set up a string spanning the entire $x$ axis, stretching from $x \rightarrow -\infty$ to $x \rightarrow \infty$. Alternatively, you can just set up a uniform scalar field throughout the entire space. Say the string moves only in the y direction and let its displacement from the $x$-axis at point $x$ and time $ct$ be $\phi(x,ct)$ (or for the field, assume it varies only along the $x$-axis). Now somehow make the string oscillate in unison across O$x$ along its entire span according to
$$
\phi(x,ct) = \phi_0 \sin(kct),\;\; \text{for}\;\text{some}\;k \in {\mathbb R}
$$
The frequency of the oscillation is simply $\omega = ck$.
If we look at this simple setup from another frame O', moving at velocity $v$ along x, the displacement of the string at location x' and time ct' will be
$$
\phi'(x',ct') = \phi_0 \sin\left( \gamma k (ct' + \beta x') \right)
$$
where $\beta = \frac{v}{c},\;\gamma = {1}{\sqrt{1-\beta^2}}$, $x' = \gamma (x-\beta ct)$, $ct' = \gamma (ct - \beta x)$ as usual, per Lorentz transforms. This is obviously a wave and it must satisfy a wave equation. So let's take a look at the usual ingredients for a wave equation:
$$
\frac{\partial \phi'}{\partial x'} = - \beta \gamma k \phi_0 cos\left( \gamma k (ct' + \beta x') \right), \;\;\frac{\partial^2 \phi'}{\partial x'^2} = - \beta^2 \gamma^2 k^2 \phi'(x',ct') \\
\frac{\partial \phi'}{\partial (ct')} = - \gamma k \phi_0 cos\left( \gamma k (ct' + \beta x') \right), \;\;\frac{\partial^2 \phi'}{\partial (ct')^2} = - \gamma^2 k^2 \phi'(x',ct')
$$
From this it follows immediately that our wave conveniently satisfies
$$
\frac{\partial^2 \phi'}{\partial x'^2} - \frac{\partial^2 \phi'}{\partial (ct')^2} = \gamma^2 k^2 (1 - \beta^2) \phi'(x',ct')
$$
or simply the Klein-Gordon equation:
$$
\frac{\partial^2 \phi'}{\partial x'^2} - \frac{\partial^2 \phi'}{\partial (ct')^2} = \frac{\omega^2}{c^2} \phi'(x',ct')
$$
What we see here is that the oscillations of the string in frame O, which are parallel to the $x$-axis, appear in O' as Klein-Gordon waves traveling along $x'$ in the direction of motion of O and at apparently faster-than-light wavefront velocities.
The reason for the apparently faster-than-light Klein-Gordon wave propagation is very simple: In O a "wavefront" corresponding to time $ct_0$ is simply a line parallel to the $x$-axis at distance $d = \phi_0 sin(kct_0)$ along $y$. All its points belong to a space-like hyperplane and are therefore not causally related. In O' the same front appears not as a wave, but as a single point propagating along $x'$ at faster-than-light "velocity" $-c/\beta$ (the negative sign arises because O' sees the string in O moving in the negative x direction). To see this, look at the coordinates of the front points in O', as given by the Lorentz transform:
$$
x' = \gamma (x-\beta ct_0) \\
ct' = \gamma (ct_0 - \beta x)
$$
Since $ct_0$ is fixed, each $x$ in O corresponds to a single $x'$ in O', which in turn corresponds to a single time $ct'$. Alternatively, eliminate $x$ or use the reverse Lorentz transform to obtain
$$
x' = - \frac{ct'}{\beta} + \frac{ct_0}{\beta\gamma}, \;\; \frac{dx'}{d(ct')} = -\frac{1}{\beta}
$$
In other words, O' observes a "propagating point" with faster-than-light velocity, but this "point" is in fact a collection of spatially-separated, causally unrelated points of the O front as observed at consecutive times in O'. Taking into account all wavefronts as generated in O, leaves O' with an observation of a "faster-than-light" Klein-Gordon wave.
And now the answer to your question :
A similar reasoning applies to a scalar field $\phi$. In O, let $\phi$ be a space-wise uniform field that oscillates in time, $\phi({\bf x}, ct) = \phi_0 \sin(\omega t)$. Its wavefront in O at a given moment $ct_0$ becomes the entire 3D-space, which is basically a constant-time, space-like hyperplane. What O' observes of this hyperplane at any moment $ct'$ in his own time is only a 2D plane, $x'=- \frac{ct'}{\beta} + \frac{ct_0}{\beta\gamma}$, perpendicular to the direction of motion of O. When O' follows this plane in time, he observes a wavefront propagating along $x'$ at faster-than-light phase velocity $-c/\beta$. The entire field appears to him as a wave $\phi'(x',ct') = \phi_0 \sin\left( \gamma k (ct' + \beta x') \right)$ satisfying the Klein-gordon equation $\Delta'\phi' - \frac{\partial^2}{\partial (ct')^2}\phi' = \frac{\omega^2}{c^2} \phi'(x',ct')$.
Conversely, we can start with a solution of the Klein-Gordon equation and boost to a frame where $\phi$ appears as a uniform field, then retrace the same reasoning.
Quantum connection:
The difference between the string Klein-Gordon eq. above and the one in relativistic quantum fields is merely the form of the coefficient on the rhs. To obtain the correct term for a field of mass m, one only has to replace the arbitrary frequency $\omega$ with the fundamental frequency (Plank frequency?) for mass m:
$$
\frac{\omega^2}{c^2} = \frac{\left(\frac{mc^2}{\hbar}\right)^2}{c^2} = \frac{m^2c^2}{\hbar^2}
$$
Everything else you know about "deriving" the Klein-Gordon in relativistic quantum theory is great, but hey, it can be done this way too.
Bottom line:
Leaving aside the string image and considering strictly scalar fields, the correct way to interpret the Klein-Gordon "faster-than-light" waves is that there exists an inertial frame where the Klein-Gordon field is uniform in space and oscillates in time at frequency $\frac{mc^2}{\hbar}$. This frame can be viewed as the "rest frame" of the particle represented by the field. The wavefronts correspond to constant time, space-like hyperplanes of the "rest frame" and necessarily "propagate" faster-than-light because this is how constant time hyperplanes transform under the Lorentz transforms. Everything else follows from the nature of space-time and there is no conflict with causality.
