Why would CuSO4-EDTA drops be used in Pb titrations?
I'm not an expert on coordination chemistry, and in particularly, metal-indicator complexes in EDTA titrations for heavy metals. But, I assume, the addition of $\ce{[Cu(EDTA)]^2-}$ solution to your unknown $\ce{Pb}$ sample is to get a sharp end-point (example of Replacement or Substitution Titration). Since, the extra metal ion added ($\ce{Cu^2+}$) has already complexed with EDTA, it won't interfere with the calculations. Meantime, both $K_{\mathrm{f}}$s of $\ce{[Cu(EDTA)]^2-}$ and $\ce{[Pb(EDTA)]^2-}$ are so closed ($5\times 10^{18}$ and $2\times 10^{18}$, respectively) that following equilibrium exists:
$$\ce{[Cu(EDTA)]^2- + Pb^2+ &lt;=&gt; [Pb(EDTA)]^2- + Cu^2+}$$
Where, $$K_{\mathrm{eq}} = \frac {[\ce{(Pb(EDTA))^2-}][\ce{Cu^2+]}}{[\ce{(Cu(EDTA))^2-}][\ce{Pb^2+]}} = \frac{K_{\mathrm{f}(\ce{(Pb(EDTA))^2-})}}{K_{\mathrm{f}(\ce{(Cu(EDTA))^2-})}} = \frac{2}{5} = 0.4$$

Thus, released $\ce{Cu^2+}$ cpmplexed with Xylenol Orange to give stable red color at operating pH (e.g., Figure 1A, Ref.2). My assumption is based on EDTA titration of $\ce{Pb^2+}$ and $\ce{Bi^3+}$  ion mixtures in the presence of Xylenol Orange as an indicator. In that titration, pH of a test sample solution containing $\ce{Pb^2+}$ and $\ce{Bi^3+}$ is adjusted to $3$  and add the indicator (Xylenol Orange). The test solution becomes red due to the complex formation of $\ce{Bi^3+}$ and Xylenol Orange (Figure 1B). When titrate with EDTA, at the first end point, the solution becomes yellow, the color of the free indicator at that pH. To titrate for $\ce{Pb^2+}$ at the same solution, it is needed to adjust pH to 5 - 5.5 until the appearance of red color again (indicating the complexation of $\ce{Pb^2+}$ with Xylenol Orange). Now, the slution can be titrated with EDTA until the turning of red to yellow. It is important to note that solution of $\ce{Pb^2+}$ ions does not become red with Xylenol Orange until pH is above 5. That means, the [$\ce{Pb^2+}$ + Xylenol Orange] complex is unstable below pH 5.



Also note that use of ammonium acetate ($\ce{NH4OAc}$) as a neutral buffer is misleading. $\ce{NH4+}$  and $\ce{AcO-}$  are not a conjugate acid/base pair, which means that they do not constitute a buffer at pH 7. However, dissolution of ammonium acetate salt in water results in pH 7, but this pH is highly labile (Ref.1). Ammonium acetate does provide buffering around pH 4.75 (the $\mathrm{p}K_{\mathrm{a}}$ of acetic acid) and around pH 9.25 (the $\mathrm{p}K_{\mathrm{a}}$ of ammonium). Thus, advantage for using $\ce{NH4OAc}$ is a drop from pH 7 to around pH 4.75, which is considerably less dramatic (e.g., addition of $\pu{1 mM} \: \ce{H+}$ to a $\pu{10 mM}$ neutral pH solution will decrease the pH to 5.8) than doing the same to pure water. Remember, during the titration, with out "real" buffer, pH of the solution changed due to release of $\ce{H+}$ ions by complexation.  

References:


L. Konermann, “Addressing a Common Misconception: Ammonium Acetate as Neutral pH “Buffer” for Native Electrospray Mass Spectrometry,” J. Am. Soc. Mass Spectrom. 2017, 28(9), 1827–1835 (https://doi.org/10.1007/s13361-017-1739-3).
Interesting reading: O. J. V. Belleza, A. J. L. Villaraza, “Ion charge density governs selectivity in the formation of metal–Xylenol Orange (M–XO) complexes,” Inorganic Chemistry Communications 2014, 47(1), 87–92 (https://doi.org/10.1016/j.inoche.2014.07.024).
Relevant: S. Murakami, “Semi-Xylenol Orange Complexes with Bivalent Metal Ions,” J. inorg, Nucl, Chem. 1981, 43(2), 335–343 (https://doi.org/10.1016/0022-1902(81)90020-8).