lo.logic - cardinal of a quotient space

Let me first treat the case where the
underlying set is infinite.

In the infinite case, your cardinal $beta$ is either $0$
or equal to $alpha$, depending on whether all points are
equivalent or not. The reason is that if the relation is
not trivial, then every point is inequivalent to some other
point, so $alphaleqbeta$, and conversely
$betaleqalphacdotalpha=alpha$ by infinite cardinal
arithmetic.

For question 1, the answer is therefore that $kappa$ is
not determined by $alpha$ and $beta$. As you observed,
$kappa$ is the number of classes, and the same infinite
set of size $alpha$ can be partitioned into any number
$kappa$ of classes, provided $1leqkappaleqalpha$.

For question 2, when $alpha$ is infinite, then since
$beta=alpha$ (unless there is only one class, in which
case $beta=0$), the bound $gamma$ is not very helpful. But the largest $kappa$ can be is $alpha$. (This is under AC; without AC, then it is possible that $kappa$ could be strictly larger than $alpha$, as I expain at the bottom.)

Similarly, for question 3, the smallest $kappa$ can be is
$1$, when $beta=0$, and otherwise, $kappa=2$ is possible,
since you can divide $alpha$ into $2$ classes, each of
size $alpha$.

In the infinite case, there are some interesting
issues that arise with the Axiom of Choice in this
question. Your observation that the quotient has size
$kappa$ seems to rely on AC, since the chains are
essentially choice functions. More generally, it was
observed in a previous MO answer by Dr. Strangechoice that
$kappa$ can actually be strictly larger than $alpha$! That is, one can partition a set into strictly more classes than there are points! For
example, consider the relation $E$ on the reals, where
$xEy$ if $x=y$ or if both $x$ and $y$ code a well order on
the natural numbers having the same order type. This is an
equivalence relation on the reals, but it is consistent
with ZF that there is no $omega_1$-sequence of reals, and
in this case there can be no injection from the $E$-classes
into the reals, since this would provide such an
$omega_1$-sequence. But there is a converse injection,
since we can injectively map reals to reals that don't code
well-orders. So this is a situation where the number of
equivalence classes is a strictly larger cardinality than
the underlying set.

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Update. In the finite case, I happened to observe that again $kappa$ is not a function of $alpha$ and $beta$. To see this, let $sim_1$ and $sim_2$ be two relations on 6 points, the first partitioning it as $2+2+2$, with three classes, and the second partitiioning it as $3+1+1+1$, with four classes. In each case, we have $alpha=6$. But unless I have made a counting mistake, it seems we also have $beta=24$ in each case, since each equivalence relation adds 3 equivalent (unordered) pairs beyond the identity pairs, making for 12 equivalent pairs (a,b), and hence $beta=36-12=24$ inequivalent pairs in each case. But the first relation has $kappa=3$ and the second has $kappa=4$; so $kappa$ is not determined by $alpha$ and $beta$.