Following @mercio's comments, I've rewritten my question in terms of zeros instead of saddles. Also, after more careful consideration, I've decided that perhaps the path I seek might not depend on the presence of poles or choice of end points, but I'm not sure. In any case, please help me prove or disprove the following statement:
In any connected region on which a meromorphic function is defined, there exists a path along which the function is smooth and analytic, and the zeros of the function and all its higher derivatives (inside the region) lie.
I was also thinking about whether such a path would be unique, but I don't need uniqueness for my problem. However, I need such a path to exist in the limit where the size of the region becomes infinite. Actually, I also need the path to be regular and differentiable, but I'll leave that question for later.
The reason I would like such a path is to consider a restriction of the function or its derivatives to the path. If the path is $z(s)=x(s)+\mathit{i} y(s)$, where s is arc length, and the function or one of its derivatives $w(z) = u(z)+\mathit{i}v(z)$, then
$$ \frac{du}{ds}=\frac{\partial u}{\partial x}x'(s)+\frac{\partial u}{\partial y}y'(s) $$ $$ \frac{dv}{ds}=\frac{\partial v}{\partial x}x'(s)+\frac{\partial v}{\partial y}y'(s) $$
are 0 iff $w'(z)=0$. Also, since $w(z)$ is continuous and finite everywhere on the path, at least 1 extremum with $\frac{du}{ds}=\frac{dv}{ds}=0$ must exist between every pair of consecutive zeros of $w(z)$. Accordingly, the extrema must coincide with where $w'(z)=0$. I could therefore conclude $w'(z)$ has at least 1 zero for each pair of consecutive $w(z)$ zeros. Moreover, the $w'(z)$ zeros would be between the corresponding $w(z)$ zeros along the path.
I'm now wondering whether this kind of reasoning is even necessary, or there is some easier way to show the result I want. I suppose it's adequate if the path just has zeros of the function and its derivative, then I can use different paths for higher derivatives. However, it would be nice if I could use the same path for all derivatives.
As an example, I've tried to search for such a path for the digamma function $\psi(z)$, which I'm studying for Convexity of reciprocal polygamma. Using Mathematica, I computed zeros of $\psi^{(m)}(z)$ for $0\leq m\leq 8$, $-1<x<0$, and $y>0$. I then looked at the path obtained from connecting zeros from $\psi^{(m)}(z)$ with increasing $m$:
The contours in the figure are labeled by the base-10 log of their values. Since the zeros can be connected in different ways, the path I obtained isn't unique.
Next, I parametrized the path connecting all zeros by arc length, $s$, then plotted $\psi^{(m)}(z)$ vs $s$:
$\psi^{(2)}(z)$ has zeros near $s=0$ and $s=0.5$. As expected, $\psi^{(3)}(z)$ has a corresponding zero near $s=0.05$ between those zeros. However, $\psi^{(2)}(z)$ also has extrema near $s=0.01$ and $s=0.35$, for which $\psi^{(3)}(z)$ has no corresponding zeros. Instead, these correspond to points where both $x'(s)=y'(s)=0$, which occurs because the path is irregular. For this path, therefore, my conclusion is invalid. It is then necessary to ask if a regular path exists, perhaps if I included more zeros, or connected the zeros in a different way?
Not an answer, but sufficient for my needs. I realized my statement that the $u(z)$ and $v(z)$ extrema along the path must coincide with where $w'(z)=0$ is false. For example, it is possible to have $\frac{du}{ds}=0$ if $\frac{\partial u}{\partial x}=0$ and $y'(s)=0$, without $\frac{\partial u}{\partial y}=0$. So even if such a path exists, and is regular and differentiable, I cannot conclude $w'(z)$ has a zero for each pair of $w(z)$ zeros in the region.
For example, consider $w(z)=z^2-1$ in some annulus around $z=0$. Then only $w(z)$ has zeros in the annulus; no derivatives do. The path $e^{\mathit{i}\theta}$ for $0\leq\theta\leq\pi$ connects the $w(z)$ zeros, and possesses extrema that do not coincide with $w'(z)=0$.
Perhaps a better question to ask is whether, in general, in the complex plane, the number of $w'(z)$ zeros is at least that of $w(z)$ less 1, but the question doesn't really serve my needs either.