$\require{AMScd} \newcommand{\RP}{\mathbb{RP}}$ Note: The splitting is actually natural in $G$ contrary to what is written in many texts. What she is referring to is the following:$\newcommand{\z}{\mathbb{Z}}$. Lets fix $G=\z/2$.
Thm: There is no possibility of a natural transformation from the functor $H_*(-,\z/2)$ to $(H_{*}(-,\z))^* \oplus Ext(H_{*-1}(-,\z),\z/2)$.
Proof: Suppose if possible that for all spaces $X,Y$ and every continuous map $X \to Y$, there is an induced a commuting square
$ \begin{CD} H_*(X,\z/2) @>\cong>> Tor(H_{*-1}(X),\z/2) \oplus H_*(X) \otimes \z/2 \\ @VVV @VVV\\ H_*(Y,\z/2) @>\cong>> Tor(H_{*-1}(Y),\z/2) \oplus H_*(Y) \otimes \z/2 \\ \end{CD}$ .
Then there would have to be a commuting square induced by $\RP^2 \to \RP^2/\RP^1 \to S^2$:
$ \begin{CD} H_2(\RP^2,\z/2) @>\cong>> Tor(H_{1}(\RP^2),\z/2) \oplus H_2(\RP^2) \otimes \z/2 \\ @| @V0VV\\ H_2(S^2,\z/2) @>\cong>> Tor(H_{2-1}(S^2),\z/2) \oplus H_2(S^2) \otimes \z/2 \\ \end{CD}$. Contradiction since the 0 map would need to be an isomorphism.
Note: One could not have expected there to be such a natural transformation because $H_*(X)$ splits into objects of different homogenous degree, while the induced maps of topological maps map elements of the same degree onto elements of the same degree.
Note 2: The diagram $ \begin{CD} H^2(\RP^2,\z/2) @>\cong>> Ext(H_{1}(\RP^2),\z/2) \oplus H_2(\RP^2)^* \\ @| @V0VV\\ H^2(S^2,\z/2) @>\cong>> Ext(H_{1}(S^2),\z/2) \oplus H_2(S^2)^* \\ \end{CD}$ shows that the exact sequence of the cohomological universal coefficient theorem does not split naturally with respect to the topological space.