$(c^{\mu}\partial_{\mu}-m)\phi(x)=0$ not Lorentz-invariant?

It depends what you mean by "breaking Lorentz invariance". As you say, it is certainly true that if you're given a vector $c^\mu$ at some point, you can construct Lorentz scalars from that vector (that is, objects that transform covariantly under Lorentz transformations). The point Zee is getting at is this: physically, a theory being Lorentz invariant means that there is no notion of a preferred reference frame. But if you specify some vector $c^\mu$ at a point, you automatically obtain a preferred reference frame constructed from it (for example, if $c^\mu$ is timelike, you can define a preferred reference frame as the one in which $c^t$ is the only nonzero component). So while something like $c^\mu \partial_\mu$ may transform covariantly, any object you construct from $c^\mu$ cannot be Lorentz-invariant because the presence of $c^\mu$ chooses a preferred frame.

(By the way, the fundamental issue here has to do with needing to specify $c^\mu$ by hand; you can remedy the situation by, for instance, upgrading $c^\mu$ to be a dynamical vector field which is not fixed a priori but is instead determined as a solution to some equations of motion.)

In general, whenever you're in doubt, a more reliable way of doing transformations is, instead of replacing things by other things, defining a new coordinate $x'^\mu = \Lambda^\mu_\nu x^\nu$ and finding equalities. In this case we have that $\partial_\mu = (\Lambda^{-1})^\nu_\mu \partial'_\nu$, so let's take the equation

$$(c^\mu \partial_\mu - m) \phi(x) = 0$$

and use the two equalities I wrote in the above paragraph:

$$(c^\mu (\Lambda^{-1})^\nu_\mu \partial'_\nu - m)\phi(\Lambda^{-1}x') = 0$$

We can use that $c^\mu (\Lambda^{-1})^\nu_\mu = \Lambda^\mu_\nu c^\nu$, so that, if we define $\phi'(x') = \phi(\Lambda^{-1}x')$, we have

$$(\Lambda^\mu_\nu c^\nu \partial'_\mu - m)\phi'(x') = 0$$

So the transformed field $\phi'$ doesn't satisfy the original equation, because you need to transform the components of $c$.