I would say there are three basic reasons for / proofs of positivity.

1. Geometry. [Kleiman 1973] proves that the number one's trying to compute is the number of points in a transverse intersection of cycles. Ergo, a nonnegative number.

2. Combinatorics. Present the cohomology ring of $Gr(k,n)$ as a quotient of that of $Gr(k,\infty)$, taking Schubert classes to Schubert classes or to zero. The latter ring is a polynomial ring containing the Schur polynomials in $x_1,\ldots,x_k$, which give the Schubert classes. There are various combinatorial proofs of the Littlewood-Richardson rule for multiplication of Schur polynomials. Once you know this rule is correct, then you know the coefficients are nonnegative.

3. Representation theory. The Schur polynomials also give the characters of "polynomial" representations of $GL(k)$, multiplication corresponds to tensor product, and decomposing in the Schur polynomial basis corresponds to decomposing the representation. Here, the coefficients are dimensions of intertwining spaces, thus nonnegative.

Perhaps the principal way to judge advantage/disadvantages of these approaches is to ask how or whether they generalize beyond the original question you ask, concerning cohomology of Grassmannians. #1 generalizes in many ways, in particular to other homogeneous spaces and other cohomology theories (equivariant, $K$, quantum, etc.) #3 generalizes to representations of other groups.

My favorite connection between #1 and #2 is Ravi Vakil's "geometric Littlewood-Richardson rule". Probably the best connection of #2 and #3 is via the theory of crystals. In both cases one can retrodict some of the combinatorial theory.

Which approach is the most insightful... I guess I have to admit a predilection for the geometry. Certainly there are many generalizations of the original problem for which we have geometric proofs of positivity but no combinatorial proofs. Maybe the simplest one involves cohomology of $d$-step flag manifolds, where $d\geq 4$ (the $d=3$ case only solved last year, in a way that doesn't extend to $d\geq 4$).

I would say there are three basic reasons for / proofs of positivity.

1. Geometry. [Kleiman 1973] proves that the number one's trying to compute is the number of points in a transverse intersection of cycles. Ergo, a nonnegative number.

2. Combinatorics. Present the cohomology ring of $Gr(k,n)$ as a quotient of that of $Gr(k,\infty)$, taking Schubert classes to Schubert classes or to zero. The latter ring is a polynomial ring containing the Schur polynomials in $x_1,\ldots,x_k$, which give the Schubert classes. There are various combinatorial proofs of the Littlewood-Richardson rule for multiplication of Schur polynomials. Once you know this rule is correct, then you know the coefficients are nonnegative.

3. Representation theory. The Schur polynomials also give the characters of "polynomial" representations of $GL(k)$, multiplication corresponds to tensor product, and decomposing in the Schur polynomial basis corresponds to decomposing the representation. Here, the coefficients are dimensions of intertwining spaces, thus nonnegative.

Perhaps the principal way to judge advantage/disadvantages of these approaches is to ask how or whether they generalize beyond the original question you ask, concerning cohomology of Grassmannians. #1 generalizes in many ways, in particular to other homogeneous spaces and other cohomology theories (equivariant, $K$, quantum, etc.) #3 generalizes to representations of other groups.

My favorite connection between #1 and #2 is Ravi Vakil's "geometric Littlewood-Richardson rule". Probably the best connection of #2 and #3 is via the theory of crystals. In both cases one can retrodict some of the combinatorial theory. The best connection I know of #1 and #3 is Belkale's construction of a tensor invariant, given an intersection point of three Schubert cycles, such that the tensor invariants constructed form a basis.

Which approach is the most insightful... I guess I have to admit a predilection for the geometry. Certainly there are many generalizations of the original problem for which we have geometric proofs of positivity but no combinatorial proofs. Maybe the simplest one involves cohomology of $d$-step flag manifolds, where $d\geq 4$ (the $d=3$ case only solved last year, in a way that doesn't extend to $d\geq 4$).