I'm reading from Milne's text https://www.jmilne.org/math/CourseNotes/RG.pdf, in chapter 8 on Lie algebras of (affine) algebraic groups $G$ over $k$. In it, he claims in 8.6 that there is an isomorphism $$ \operatorname{Hom}_{k-\text{linear}}(\ker(\epsilon)/\ker(\epsilon)^2, k) \to \operatorname{Tgt}_e(G), $$ where $\epsilon$ is the coidentity map and $\operatorname{Tgt}_e(G) = \ker(G(k[\varepsilon]) \to G(k))$. I understand why there is such a bijection: an element of $\operatorname{Tgt}_e(G)$ is a morphism $\mathcal O(G) \to k[\varepsilon]$ whose composite with $k[\varepsilon] \to k$ equals $\epsilon$, from which you see it factors through $\mathcal O(G)/\ker(\epsilon)^2$. Then using $\mathcal O(G)/\ker(\epsilon)^2 = k \oplus \ker(\epsilon)/\ker(\epsilon)^2$ you see it is completely determined by a linear map $\ker(\epsilon)/\ker(\epsilon)^2 \to k$.
But why is it a group homomorphism? The group structure on $\operatorname{Tgt}_e(G)$, viewing elements of it as morphisms $\mathcal O(G) \to k[\varepsilon]$, is determined by the comultiplication map $\Delta \colon \mathcal O(G) \otimes_k \mathcal O(G) \to \mathcal O(G)$. I do not understand how this agrees with pointwise addition on the set of linear maps, and my attempts to prove it were not fruitful.