How do algebraists intuitively picture normal subgroups and ideals?

Normal subgroups and ideals are kernels. That is the reason they are interesting, they are precisely the subgroups/rings that are kernels of homomorphisms.

By themselves, ideals and normal subgroups are typically not interesting. They become interesting when considered together with the homomorphisms of which they are the kernels: that is, we are interested in the structure of $G/N$ or $R/I$, and how they, together with the structure of $N$ and $I$, give information about the structure of $G$ and $R$.

To add to existing answers, ideals in rings are useful beyond being kernels. For example, ideals in a ring $R$ are $R$-modules (abelian groups with an action of $R$) and sometimes have an interesting structure in that sense.

Furthermore, in algebraic geometry we treat the set of all prime ideals in a commutative ring as a topological space with additional structure; this is called the spectrum of the ring. Upon being told it's hard to imagine how this could be a useful topological space; the original motivation is that the set of all polynomials (in any number of dimensions) that vanish on a particular set form a prime ideal. Thus prime ideals could also be seen as points in this sense.

To elaborate on a very important point briefly touched on in Matt Samuel’s answer:

• in algebraic geometry, a ring $R$ is viewed as representing a space $X$, with elements of the ring being some kind of functions on the space — extending the way that e.g. the polynomial ring $\mathbb{Z}[x,y]$ can be viewed as a ring of functions on the 2-dimensional plane.

• any subspace $X' \subset X$ of the space should then induce an ideal $I_X \subseteq R$, the ideal of functions that vanish (i.e. are constantly 0) on the subspace $X'$. So an ideal can represent a subspace; for instance, the ideal $(x-1) \subseteq \mathbb{Z}[x,y]$ represents the line $x = 1$ in the plane, since a polynomial vanishes on that subspace exactly if it’s a multiple of $x-1$.

• slightly more generally, the functions that “vanish to higher order” on a subspace also form an ideal. E.g. $((x-1)^2)$ represent the functions that vanish to second-order on the line $x=1$. If you haven’t met this idea of higher-order vanishing before, look at (or imagine) a 3-d graph of the function $z = (x-1)^2$ on the $xy$-plane, and see how it’s not just zero on the line $x=1$, it’s flat on that line. Intuitively, one can think of it as vanishing not just on that line, but on an infinitessimal thickening of that line. (And $(x-1)^3$ vanishes on a slightly thicker infinitessimal thickening, and so on.)

So intuitively, in the algebraic geometry picture, ideals of a ring correspond to subspaces of a space.