The desired relation can be written in components as
$$f_n(\mathbf{z}^{(1)},\mathbf{\bar z}^{(1)})-f_n(\mathbf{z}^{(2)},\mathbf{\bar z}^{(2)})=\int_0^1 d\tau\,\sum_{m}\left(\frac{\partial f_n}{\partial z_m}(z_m^{(1)}-z_m^{(2)})+\frac{\partial f_n}{\partial \bar{z}_m}(\bar{z}_m^{(1)}-\bar{z}_m^{(2)})\right),\;\;[1]$$
with the prescribed $\tau$ dependence:
$$\mathbf{z}(\tau)=\mathbf{z}^{(2)}+\tau(\mathbf{z}^{(1)}-\mathbf{z}^{(2)}),\;\;
\mathbf{\bar z}(\tau)=\mathbf{\bar z}^{(2)}+\tau(\mathbf{\bar z}^{(1)}-\mathbf{\bar z}^{(2)}).\;\;[2]$$
We start from the equation
$$f_n(\mathbf{z}^{(1)},\mathbf{\bar z}^{(1)})-f_n(\mathbf{z}^{(2)},\mathbf{\bar z}^{(2)})=\int_0^1 d\tau\,\frac{d}{d\tau}f_n(\mathbf{z},\mathbf{\bar z}).\;\;[3]$$
Equation [3] implies equation [1] if
$$df_n=\sum_m\left(\frac{\partial f_n}{\partial z_m}dz_m+\frac{\partial f_n}{\partial \bar{z}_m}d\bar{z}_m\right).\;\;[4]$$
This is indeed a property of the Wirtinger differential.$^\ast$

$^\ast$ **Proof**

Denote $z=x+iy$, $\bar{z}=x-iy$ and insert the definitions $\partial/\partial x=\partial/\partial z+\partial/\partial \bar{z}$, $\partial/\partial y=i\partial/\partial z-i\partial/\partial \bar{z}$ of the Wirtinger derivatives:

$$df_n=\sum_m\left(\frac{\partial f_n}{\partial x_m}dx_m+\frac{\partial f_n}{\partial y_m}dy_m\right)$$
$$\qquad=\sum_m\left(\frac{\partial f_n}{\partial z_m}dx_m+\frac{\partial f_n}{\partial \bar{z}_m}dx_m+i\frac{\partial f_n}{\partial z_m}dy_m-i\frac{\partial f_n}{\partial \bar{z}_m}dy_m\right)$$
$$\qquad=\sum_m\left(\frac{\partial f_n}{\partial z_m}dz_m+\frac{\partial f_n}{\partial \bar{z}_m}d\bar{z}_m\right).$$