There isn't a formula for what you're looking for. At least, the formula can't make sense for all initial data and depend continuously on that initial data.
Your initial data is two points in the Stiefel manifold $V_{n,d}$, together with two points in $\mathbb R^n$. If you were to find an embedding $B^d \to \mathbb R^n$ whose derivatives agreed at two points with these points of your Stiefel space, you'd have produced a section of the fibration:
$$ Emb(B^d, \mathbb R^n) \to V_{n,d}^2 \times C_2 \mathbb R^n$$
where this map is given by restriction to the endpoints. $C_2 \mathbb R^n$ is the configuration space of two points in $\mathbb R^n$, it has the homotopy-type of a sphere.
$Emb(B^d, \mathbb R^n)$ has the homotopy-type of $V_{n,d}$. So if your section existed, the homotopy and homology groups of $V_{n,d}^2 \times C_2 \mathbb R^n$ would embed in the homotopy and homology groups of $V_{n,d}$. Since the first non-trivial homotopy group of $v_{n,d}$$V_{n,d}$ is cyclic, this is impossible.
Admittedly, I'm making some extra assumptions that you did not specify, but I think this shows you that whatever your solution is, it has to be somewhat nuanced.