A linear algebraic group $G$is an almost direct product of its subgroups $G_1,\dots,G_r$ if the product map $$ G_1\times\dots\times G_r\to G:(g_1,\dots,g_r)\mapsto g_1\cdots g_r$$ is a surjective morphism with finite kernel (also known as an isogeny).
Let $N=\{(I,I),(-I,-I)\}$ and $G=(\operatorname{SL}_2\times\operatorname{SL}_2)/N$. Then $G$ is an almost direct product of $\operatorname{SL}_2$ and $\operatorname{SL}_2$, but not a direct product of almost simple subgroups.
How can I show this? We need subgroups of $G$ which can be identified with $\operatorname{SL}_2$. Such a subgroup is of the form $H/N$ where $H\le \operatorname{SL}_2\times\operatorname{SL}_2$ and $N\subseteq H$. How do I define the product map? The expected product map $H/N\times H/N\to G: (h_1N,h_2N)\mapsto h_1h_2N$ does not work since elements in $G$ would be of the form $(a,b)N$ with $a,b\in\operatorname{SL}_2$.