Metric on Steifel and Grassmannian manifolds generalizing Fubini-Study

There is no need to proceed via the Stiefel-Manifolds. You can directly realize the Grassmannians as homogeneous spaces. In the real case, this takes the form $G(k,n)=SO(n)/S(O(k)\times O(n-k))$. This corresponds to a so-called symmetric decomposition of the Lie algebra $\mathfrak{so}(n)$, thus making $G(k,n)$ into a compact symmetric space. (The Riemannian metric is induced from the restriction of the Killing form of $\mathfrak{so}(n)$ to the orthocomplement of $\mathfrak{o}(k)\times\mathfrak{o}(n-k)$). The resulting metric is relatively simple, but does not have constant curvature. There is a lot known about such spaces, "Riemannian symmetric spaces" is the right key-word to start looking for. The picture for the complex and quaternionic fields is quite similar, you just have to replace $SO$ by $SU$ respectively $Sp$ (quaternionically unitary group).

It is an standard fact that the space of real grassmanian $G(k,n)$ is homeomorphic to all projections in $M_{n}(\mathbb{R})$ with trace $k$.(A projection is a matrix with $A=A^{tr}=A^{2}$. The inner product of $M_{n}(\mathbb{R})\simeq \mathbb{R}^{n^{2}}$ is $trace(AB^{tr})$

This inner product is invariant under the action of $O(n)$. So actually O(n) is acting on G(k,n) isometricaaly and transitively. The action is transitive since every two projections with the same rank are unitary equivalent, a standard fact in linear algebra. It is also mentioned in "k theory and C* algebra, A freilndly approach" By Wegg Olsen. this would implies that the metric has constant curvature.