Geometrical interpretation of $I(X_1\cap X_2)\neq I(X_1)+I(X_2)$, $X_i$ algebraic sets in $\mathbb{A}^n$

First let me congratulate you: your intuition is remarkably correct and your example is excellent.

In the land of scheme theory things are astonishingly simple:
Given two subschemes $X_1= V(I_1), X_2=V(I_2)\subset \mathbb A^n _k$, their intersection $X_1\cap X_2=V(I_1+ I_2)$ is defined by the sum of their ideals, period .
In your example $I_1=(x), \: I_2=(x+y^2) $ and so the intersection $X_1\cap X_2$ is given by the ideal $I_1+ I_2=(x, y^2)$.
This is carefully to be distinguished from the subscheme $$(X_1\cap X_2)_{red}=V(\sqrt{I_1+ I_2})= V(x,y)\subsetneq X_1\cap X_2 $$ which is smaller in the sense that it has the same underlying set (the origin $(0,0)$), but less functions living on it: only the elements of $k$, whereas on the genuine intersection $X_1\cap X_2$ the set of regular functions is the non reduced ring $k[y]/(y^2)$.
The non reducedness is the translation of the tangency of $X_1$ and $X_2$ at $(0,0)$.

In the more barren land of classical varieties you are constrained to reduced ideals and their corresponding varieties.
Unfortunately this implies that even if you want to intersect two reduced varieties, the intersection will not be reduced, as in the example examined above, and you will be forced to reduce that intersection by taking the root $\sqrt{I_1+ I_2}$ of the genuine ideal $I_1+ I_2$ of the intersection.

Edit
I'll add a few words about your last question, the difference between $(x,y^2)$ and $(y,x^2)$, which I forgot to address.
The first ideal can be seen as that of the intersection $S=V\cap L=V(x,y^2)$ of the parabola $P=V(x-y^2)$ and the vertical line $L=V(x)$.
From the classical varieties point of view this is just the origin $O=(0,0)$.
But scheme-theoretically you add the information that the ring of regular functions is $k[x,y]/(x, y^2)=k[y]/( y^2)$, so that a function on $S$ is of the form $q+r\bar y$.
In other words, the intersection $S$ is a little bigger than just$O$, in that if you restrict a polynomial $F(x,y)=a+bx+cy+dx^2+...$ to $S$, you will get $a+c\bar y$ :you can compute $c=\frac {\partial F}{\partial y}(0,0)$ for a function on $S$!
A similar analysis applies to $T=V(y,x^2)$ on which you can compute $\frac {\partial F}{\partial x}(0,0)$