Union of closed subschemes with the structure sheaf over it

As with all definitions, there is no "proof" that the adopted definition is the right one but only a feeling that it better corresponds to our intuition.

In the case at hand, taking $R/IJ$ as structure sheaf would make union schemes nonreduced for no good reason. For example take $R=k[x,y,z], I=(y,z), J=(x,z)$. Geometrically you are describing the union $U$ of the $x$-axis and the $y$-axis in affine three space $\mathbb A^3_k$ . It should have a reduced structure, correctly described by $I\cap J$, whereas $I.J=(xy,zx, zy, z^2)$ would make the function $z$ nilpotent but not zero on $U$, which feels wrong since $U$ should be a closed subscheme of the plane $z=0$.

A more brutal objection to the idea of defining the union of two subschemes by the product of their ideals is that a subscheme $U$ would practically never be equal to its union with itself, since in general$ I\neq I^2$ : we would have (almost always) $$U\neq U \cup U$$
That looks bad!

(i) The union of two closed subschemes should be the smallest closed subscheme containing the two given ones; if you like, an initial object in the category of all closed subschemes containing the given two. In the case when the two given closed subschemes are $V(I)$ and $V(J)$, this is is $V(I\cap J)$. (The operation $V$ interchanges closed sets and ideals, and is order reversing.)

(ii) In function-theoretic terms, a function $f$ (i.e. an element of the ring $A$) should vanish on $V(I) \cup V(J)$ (i.e. lie in the ideal cutting out $V(I)\cup V(J)$) if and only it vanishes on both $V(I)$ and $V(J)$ (i.e. lies in $I$ and $J$), which happens if and only if $f$ lies in $I\cap J$. Thus we are forced to define $V(I)\cup V(J) = V(I\cap J)$, if union is to have anything like its usual meaning.

Because the first one is the right answer in the case of affine varieties, and the second one is not. Indeed, $R/I$, $R/J$ are nilpotent-free implies $R/I\cap J$ is nilpotent-free, but not so for $R/IJ$.