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Let us write the Wilson-line of a simple open curve $\gamma: [s_i,s_f]\to \mathbb{R}^4$ as $$ U(s_f,s_i) ~=~ \mathcal{P}\exp \left[ i\int_{\gamma} A_{\mu}~ dx^{\mu} \right].\tag{1} $$

The path-ordering $\mathcal{P}$ becomes important if the gauge potential $$A_{\mu}~=~A^a_{\mu} T_a\tag{2}$$ is non-abelian. Here $T_a$ are the generators of the corresponding Lie algebra.

The Wilson-line has groupoid properties, e.g., $$U(s_3,s_2)U(s_2,s_1)~=~ U(s_3,s_1), \qquad U(s,s) ~=~ {\bf 1}.\tag{3}$$

If one differentiates wrt. the final point $s_f$, one gets $$\frac {\mathrm{d}U(s_f,s_i)}{\mathrm{d}s_f} ~=~ i\dot{\gamma}^{\mu}(s_f)~A_{\mu}(\gamma(s_f)) ~U(s_f,s_i). \tag{4}$$

If one differentiates wrt. the initial point $s_i$, one gets $$ \frac {\mathrm{d}U(s_f,s_i)}{\mathrm{d}s_i} ~=~ -U(s_f,s_i)~i\dot{\gamma}^{\mu}(s_i)~A_{\mu}(\gamma(s_i)) . \tag{5}$$

OP wants to differentiate the Wilson-line $U(s_f,s_i)$ functionally wrt. the gauge potential components $A^a_{\mu}(x)$. One gets $$ \frac {\delta U(s_f,s_i)}{\delta A^a_{\mu}(x)} ~=~\int_{s_i}^{s_f}\! \mathrm{d}s~ U(s_f,s)~ i\dot{\gamma}^{\mu}(s)\delta^4(x-\gamma(s))T_a~U(s,s_i). \tag{6}\label{eq:6}$$

Heuristic proof of $\eqref{eq:6}$. Since we have already used the letter $x\in\mathbb{R}^4$ in $\eqref{eq:6}$ as a fixed space-time point, let us call an arbitrary spacetime point for $y\in\mathbb{R}^4$.

• Imagine that $\tilde{A}(y)=A(y)+\delta A(y)$ is an infinitesimal variation of the gauge potential $A(y)$.

• Imagine that $\delta A(y)$ only differs from zero in an infinitesimally small neighborhood $\Omega$ of the fixed space-time point $x$.

• Assume that the curve $\gamma$ intersects the neighborhood $\Omega$ at the parametervalue interval $[s_x-\varepsilon,s_x+\varepsilon]\subseteq [s_i,s_f]$. (If the curve $\gamma$ does not intersect the neighborhood $\Omega$, then the equation $\eqref{eq:6}$ becomes trivially correct: $0=0$.)

On one hand, such infinitesimal variation of the gauge potential yields $$\delta U(s_f,s_i)~=~U(s_f,s_x+\varepsilon)~\delta U(s_x+\varepsilon,s_x-\varepsilon)~U(s_x-\varepsilon,s_i), \tag{7}\label{eq:7}$$ and $$\begin{align}\delta U(s_x+\varepsilon,s_x-\varepsilon)~\approx~&2\varepsilon i~ \dot{\gamma}^{\mu}(s_x)~\delta A_{\mu}(\gamma(s_x)) \cr ~=~&\int_{\Omega} \!\mathrm{d}^4y~\delta^4(y-\gamma(s_x))~2\varepsilon i\dot{\gamma}^{\mu}(s_x)~\delta A_{\mu}(y)\cr ~\approx~& \int_{\Omega} \!\mathrm{d}^4y~\int_{s_x-\varepsilon}^{s_x+\varepsilon}\! \mathrm{d}s~\delta^4(y-\gamma(s))~i\dot{\gamma}^{\mu}(s)~\delta A_{\mu}(y).\end{align}\tag{8}\label{eq:8}$$ On the other hand, the defining property of a functional derivative yields $$\begin{align}\delta U(s_f,s_i) ~=~&\int_{\mathbb{R}^4} \!\mathrm{d}^4y~ \frac {\delta U(s_f,s_i)}{\delta A^a_{\mu}(y)} ~\delta A^a_{\mu}(y)\cr &~=~\int_{\Omega} \!\mathrm{d}^4y~ \frac {\delta U(s_f,s_i)}{\delta A^a_{\mu}(y)} ~\delta A^a_{\mu}(y).\end{align}\tag{9}\label{eq:9}$$ A comparison of eqs. $\eqref{eq:7}$, $\eqref{eq:8}$ and $\eqref{eq:9}$ yields eq. $\eqref{eq:6}$.