= Solution
Take $G=S_3$ and its <normal subgroup> $N=A_3=\langle r\rangle$, where $r=(1\ 2\ 3)$. Normality follows either from the <sign of a permutation> or directly from conjugation sending $r$ to $r$ or $r^{-1}$. The <quotient group> has order two, and $t=(1\ 2)$ provides a section $i:C_2\to S_3$ because $t^2=1$ and $tN$ is its nonidentity coset.
However, the map $\Phi(q,n)=i(q)n$ from the <direct product of groups> is not a <group homomorphism>. In its domain $(1,r)$ commutes with $(tN,1)$, whereas their images $r$ and $t$ do not commute: $trt^{-1}=r^{-1}\ne r$. This is a <split group extension> giving a genuine <semidirect product> instead of a direct product. \b[The pair $S_3,A_3$ supplies the required split but non-direct example.]
For the general bijectivity assertion, let $\pi:G\to G/N$ be the quotient projection and assume $\pi\circ i$ is the identity. Given $g\in G$, put $q=\pi(g)$ and $n=i(q)^{-1}g$. Then $\pi(n)=q^{-1}q=1$, so $n\in N$ and $g=i(q)n$. This proves surjectivity of $\Phi$. If $i(q)n=i(q')n'$, applying $\pi$ gives $q=q'$, after which cancellation gives $n=n'$. Thus
$$
\boxed{\Phi:(G/N)\times N\longrightarrow G\text{ is always a bijection.}}
$$
This <section normal form for a split group extension> does not require finiteness. The obstruction to its being a homomorphism is the nontrivial conjugation action of the section on $N$.
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