OP Here, I've figured out the issue. It really is a functional derivative. Below I include an excerpt from my own notes, in case another lost soul stumbles upon this page in the future. (Please excuse the changes in notation from my original post, I hope it won't be difficult to switch over, and please excuse the fact that some of this might be in excruciating detail - also certainly not the most efficient way of deriving the result)
$\newcommand{\D}{\partial}$ $\newcommand{\dd}{d}$
We've already encountered <PERSON> lines through a <PERSON> diagram analysis. It will turn out that they are a common object that arises when computing the propagator in the presence of a background field. In the most general form, <PERSON> lines are:
\begin{align} V(z,y) \equiv \hat{P} \exp \left(ig \int_{y}^{z} A_\mu(s)\dd s^\mu\right). \end{align}
$\hat{P}$ is the path ordering operator. The integral here is a line integral in spacetime from some starting point $y^\mu$ to some ending point $z^\mu$, along some path that we can parametrize via $s^\mu(t)$.
\begin{align} V(z,y) = \hat{P} \exp \left( ig \int_0^1 A_\mu(s(t)) \frac{ \dd s^\mu}{\dd t} \dd t\right). \end{align}
We've defined $s^\mu(0) = y^\mu$ and $s^\mu(1) = z^\mu$. In order to see how this might appear as a solution to QCD equations of motion, we'll need to know how to take derivatives of <PERSON> lines. 818
We assume here that the background field $A_\mu$ is a fixed function, and therefore the <PERSON> line is a functional of the path $s^\mu(t)$. We want to take a derivative with respect to spacetime coordinates $\frac{ \D }{\D x^\mu}$. Since $V(z,y)$ is a functional, we can think of it as a function of an infinite number of variables $s_i^\mu$. The label $i$ is purely symbolic, since it's really a function of $s(t)$ for every value of $t$. Using this symbolic notation, one can then use the chain rule:
\begin{align} \frac{ \D V(z,y)}{\D x^\mu} = \frac{ \D V(z,y)}{\D s_i^\nu}\frac{ \D s_i^\nu}{\D x^\mu}. \end{align}
Moving back to the functional notation, we can write $s_i^\nu = s^\nu(t)$ and then the implied sum over $i$ becomes an integral over $t$.
\begin{align} \label{Dform} \frac{ \D V(z,y)}{\D x^\mu} = \int_0^1 \dd t \frac{ \delta V(z,y)}{\delta s^\nu(t)} \frac{ \D s^\nu}{\D x^\mu}. \end{align}
Now it is clear that we cannot go further without figuring out the functional derivative of the <PERSON> line with respect to the path. Let's recall the method of finding the functional derivative. First, let's write $V(z,y)$ as a functional of $s^\nu(t)$.
\begin{align} V[s(t)] = \hat{P} \exp\left( ig\int_0^1 \dd t A_\mu(s(t))\frac{ \dd s^\mu}{\dd t}\right). \end{align}
Then let's make the transformation $s^\mu(t) \rightarrow s^\nu(t) + \delta s^\mu(t)$. We'll define $\delta V = V(s+\delta s) - V(s)$. 818