At least when working theoretically, or by hand, it is probably easiest to solve this sort of question locally. (I don't have any feeling for how computationally implemented algorithms for this sort of question work, and so won't comment any more on them.)
In short, fix a prime $v$ of $K$, and consider the completion $K_v$ and the local
extension $K_v[\sqrt{d}]/K_v$. The discriminant of $K[\sqrt{d}]/K$ will be a product of the local discriminants, and so we are reduced to computing these.
How do we analyze $K_v[\sqrt{d}]/K_v$? Well, you could first try to compute
the finite index subgroup $(K_v^{\times})^2$ of $K_v^{\times}$. Note that
$K_v^{\times} = \mathcal O_{K_v}^{\times} \times \pi^{\mathbb Z}$, if $\pi$
is some choice of uniformizer, and so $(K_v^{\times})^2 =
(\mathcal O_{K_v}^{\times})^2\times \pi^{2\mathbb Z}$.
Computing $(\mathcal O_{K_v}^{\times})^2\subset \mathcal O_{K_v}^{\times}$
is straightforward. E.g. if $v$ has odd residue characteristic, then
any element in $1 + \pi \mathcal O_{K_v}$ is a square, and so it is just
a question of computing the subgroup of squares in the residue field at $v$.
If $v$ has even residue characteristic, then it is still pretty easy in any
particular case to compute the square units, say using the fact that
any unit congruent to $1 \bmod 4\pi$ is a square (as one sees using
the binomial expansion for $(1+x)^{1/2}$).
Once you've done this step, you can easily check if $d$ is a square in $K_v$,
which tells you whether or not $K[\sqrt{d}]/K$ is split at $v$.
To compute the discriminant is not that much harder. If $v$ has odd residue characteristic,
then we may divide $d$ through by even powers of $\pi$ so that it is either
a unit, or else is exactly divisible by $\pi$. In the first case, $K_v[\sqrt{d}]/K_v$ will be unramified at $v$, and hence the contribution to
the discriminant from $v$ will be $1$. In the second case, $K_v[\sqrt{d}]/K_v$ will be tamely ramified at $v$, and the contribution to the discriminant from $v$ will be one power of $v$.
If $v$ is of even residue characteristic, then the computations are more involved, because even if $d$ (perhaps after dividing through by even powers of
$\pi$) is a unit, it can still happen that $K_v[\sqrt{d}]/K_v$ is ramified
(consider the case $\mathbb Q_2(i)/\mathbb Q_2$), although it need not be
(consider the case $\mathbb Q_2(\sqrt{5})/\mathbb Q_2$).
If you would like more details, I can provide them.
Best Answer
As far as I know there is no way of finding the discriminant without finding the full ring of integers in the process.
Indeed, once you know the ring of integers, finding the discriminant is a trivial piece of linear algebra; and conversely, if you know the discriminant in advance, that makes finding the ring of integers much easier (because once you've found enough integers to generate a subring with the right discriminant, you know you can stop). So determining the discriminant cannot be all that much easier than determining the integers.
As for how to find the integers, it is a very well-studied problem; if you're interested in such things, Henri Cohen's "A course in computational algebraic number theory" is a very good reference.