In a coalgebra $A$ with comultiplication $\Delta$ and counit $\epsilon$, a two-sided coideal is defined to be a linear subspace satisfying $\Delta (I) \subseteq A\otimes I + I \otimes A$ and $I \subseteq \operatorname{ker} (\epsilon)$.
How do I understand this as the categorical dual notion of an ideal of an algebra (or ring), which is defined by $rI \subseteq I$ and $Ir \subseteq I$ for $r \in A$? I am particularly confused why there is an "additional" kernel condition for a coideal, which does not seem present in the definition of an ideal.
The $\Delta(I) \subseteq A\otimes I + I\otimes A$ part is easier, so let's begin with that! If we have a ring $R$ and we denote its multiplication by $m\colon R\otimes R\to R$, then any two-sided ideal $I$ of $R$ satisfies both $m(R\otimes I)\subseteq I$ and $m(I\otimes R)\subseteq I$. It then follows that $m(R\otimes I + I\otimes R)\subseteq I$. The "reverse all arrows in the diagrams" logic then also flips the subset sign, and this provides the first condition.
The second condition boils down to the idea that we want (co)ideals to be such that the corresponding quotients are (co)algebras. Furthermore, we want the projection $p\colon C\to C/I$ to be a natural morphism of (co)algebras. For this to be true, we need $\varepsilon_C = \varepsilon_{C/I}\circ p$. For the sake of argument, let's suppose that there is an element $c\in I$ but $c\not\in\ker(\varepsilon_C)$. Then, by definition, $\varepsilon_C(c)\neq0$ but $(\varepsilon_{C/I}\circ p)(c) = \varepsilon_{C/I}(0) = 0$, causing a problem!
I hope this helps :)