Consider the manifold $M := \operatorname{SO}(n) \times \mathbb{S}^{n-1}$, endowed with the product metric given by the bi-invariant metric of $\operatorname{SO}(n)$ and the round metric of $\mathbb{S}^{n-1}$.
Consider the left action of $\operatorname{SO}(m)$, for some $m \leq n$ on $M$ given as $(g, (x, y)) \mapsto (xg^t, gy)$ for every $g \in \operatorname{SO}(m)$ and every $(x, y) \in \operatorname{SO}(n) \times \mathbb{S}^{n-1}$. Seeing $\operatorname{SO}(m)$ as a subgroup of $\operatorname{SO}(n)$ of the form $$ \begin{pmatrix} I_{n-m} & 0\\ 0 & U \end{pmatrix}, $$ with $I_{n-m}$ the identity matrix of dimension $n-m$ and $U \in \operatorname{SO}(m)$.
Since the action is free, smooth and proper, there exists a metric on $M/\operatorname{SO}(m)$ such that the map $\pi: M \to M/\operatorname{SO}(m)$ is a Riemannian submersion with totally geodesic fibres.
From O'Neill's formula, we know that for every pair of orthonormal horizontal vectors $U, W \in T_{(x,y)} M$, $$ K_{M/\operatorname{SO}(m)}(\pi_* U, \pi_* W) = K_M(U, W) + \frac{3}{4}|[U, W]^V|^2, $$ where $K_{M/\operatorname{SO}(m)}$ is the sectional curvature of $M/\operatorname{SO}(m)$, $K_M$ is the sectional curvature of $M$, and $[U, W]^V$ is the vertical component of the Lie derivative of $U$ and $W$. (Note that the latter is well-defined since for every two horizontal vector fields $\tilde{U}$, $\tilde{W}$ extending $U, W$, the vertical component of their Lie derivative $[\tilde{U}, \tilde{W}]^V$ only depends on $U$ and $W$).
My question is: is it possible to upper bound $|[U, W]^V|^2$? This way, it would be possible to upper bound the curvature of the quotient space.
Update: I am not sure if it would be helpful to see $\mathbb{S}^{n-1}$ as the quotient $\operatorname{SO}(n)/\operatorname{SO}(n-1)$, that way one can see the total space $M$ as a Lie group ($\operatorname{SO}(n) \times \operatorname{SO}(n)$) endowed with the bi-invariant metric. Modifying the action accordingly, which now would be given by the group $\operatorname{SO}(m) \times \operatorname{SO}(n-1)$, as $$ ((g, h), (x, y)) \mapsto (xg^t, gyh), $$ for any $(g, h) \in \operatorname{SO}(m) \times \operatorname{SO}(n-1)$, and any $(x, y) \in \operatorname{SO}(n) \times \operatorname{SO}(n)$. In this setting, I believe the quotient space is a bi-quotient (see @RamiroLafuente 's comment). Does this make things any simpler or clearer?
Clarification: I know there exists a bound from the fact that the quotient space is compact, but I need something explicit. At least I would need how the curvature grows (linearly, polynomially, exponentially...) with respect to the dimensions of the original manifold and the group acting on it.
