Let $K$ be a field and $G\subset K^m$ an (affine) algebraic group.
If $\varphi:G\rightarrow (K^n,+)$ is a morphism of algebraic groups, why is $\varphi(G)$ is an algebraic group ?
I would say for instance that $\varphi(G)$ is constructible by Chevalleys Theorem (if $K$ is algebraically closed, isn't it ?), and a group so closed... hence an algebraic variety, but that seems to involve to much technicalities, and does not hold if $K$ is not algebraically closed (does it ?). A more direct way to see this ?
First, Chevalley's theorem holds pretty universally—it holds for any morphism locally of finite presentation (or maybe you need actual finite presentation, I can't remember).
Anyways, the true reason is fairly sophisticated if you don't assume that your groups are smooth. The reference is then SGA 3 Proposition 1.2, Exposé VIB.
If your groups ARE smooth (which, for example, is always true in characteristic $0$), then, in fact any morphism $f:G\to H$ is a quotient map onto the scheme-theoretic image which is a smooth group scheme itself. The fact about it being a quotient map and smooth is just basic theory. The hard part is the Closed Orbit Lemma. See here for a (not too rigorous) discussion of the topic.