The reason to care about the discriminant of a polynomial $f(x) \in K[x]$ as a way to detect double roots of $f(x)$ in some extension of $K$ is that it is a number whose vanishing determines whether or not $f(x)$ has a repeated root and this number lies in $K$. So without leaving $K$, we can make a calculation about $f(x)$ that tells us whether or not $f(x)$ has a repeated root. (That's why we don't use, say, $\prod_{i < j} (r_j - r_i)$ without squaring as the discriminant.) This is analogous to the utility of being able to tell whether or not $f(x) \in K[x]$ has a repeated root in some extension of $K$ by checking if $f(x)$ and $f'(x)$ in $K[x]$ are relatively prime.
It turns out that the discriminant has more uses, but this is not to be expected in advance. In this way the discriminant is like many other concepts in math whose applications go far beyond their original purpose over time:
when ${\rm disc}(f)$ is nonzero, whether or not it is a square in $K^\times$ is related to whether or not the action of the Galois group of $f$ over $K$ on the $n$ roots of $f(x)$ has image in $A_n$ or not, which when $n = 3$ lets us compute the splitting field of $f$ over $K$ by knowing the square class of ${\rm disc}(f)$ in $K^\times/(K^\times)^2$,
the interpretation of a polynomial discriminant as the determinant of a matrix whose components describe a trace-pairing generalizes to the discriminant of a quadratic form $Q(x_1,\ldots,x_n)$ with coefficients in a field $K$ as the determinant of a matrix whose components describe a bilinear-pairing associated to $Q$, and when ${\rm disc}(Q) \not= 0$ the square class that ${\rm disc}(Q)$ belongs to in $K^\times/(K^\times)^2$ can help us show quadratic forms over certain fields are inequivalent,
the appearance of the discriminants in several places in algebraic number theory links it to computing rings of integers, to giving us an algebraic analogue of the geometric idea of volume (see here), to ramification of primes, and to determining the extent to which a ring of integers fails to be self-dual (see here).
Over a finite field $k$ of odd characteristic, the discriminant of a separable polynomial $f(x)$ in $k[x]$ is related to the parity of the number of irreducible factors of $f(x)$ in $k[x]$: $\chi({\rm disc}(f)) = (-1)^{n-d}$, where $\chi$ is the unique quadratic character on $k^\times$, $n = \deg f$ and $d$ is the number of monic irreducible factors of $f(x)$ in $k[x]$. Equivalently, the discriminant helps us compute the Moebius function on $k[x]$ without having to factor $f(x)$:
$$
\mu(f) = (-1)^d = (-1)^n \chi({\rm disc}(f)).
$$
In particular, if we can compute that $\mu(f) = 1$ then we have proved $f(x)$ is reducible since irreducibles have Moebius value $1$. The analogue of this when $k$ has characteristic $2$ (so $k^\times$ has no quadratic character) is more subtle. The case $k = \mathbf Z/(2)$ is explained and put to use in my answer to a MSE question here.
