Realizing possible accelerations of paths on a sphere $\newcommand{\al}{\alpha}$
Let $x,v \in \mathbb{S}^n \subseteq \mathbb{R}^{n+1}$, $w \in \mathbb{R}^{n+1}$ satisfy
$\langle x,v \rangle=0, \langle x,w \rangle=-1$.
Does there exist a smooth path $\alpha:(-\epsilon,\epsilon) \to \mathbb{S}^n$ such that
$$
\alpha(0)=x, \dot \alpha(0)=v, \ddot \alpha(0)=w?
$$

The condition $\langle x,w \rangle=-1$ is necessary for the existence of such a path:
Differentiating twice $\langle \alpha,\alpha \rangle=1$ we get
$$ 
\langle \dot \alpha,\dot \alpha\rangle+\langle \alpha,\ddot\alpha \rangle=0.
$$
I tried to play with paths of the form of
$$
\al(t)=a(t)x+b(t)v(t),
$$
where $v(t) \in T_x\mathbb{S}^n$, but so far my computations did not reach a definite conclusion.
Note that every path $\al$ can be written in the above form. Part of the problem here is that $a(0)=1$, so differentiating $b(t)=\sqrt{1-a^2(t)}$ is not trivial at $t=0$.
 A: $\newcommand{\brak}[1]{\left\langle#1\right\rangle}$tl; dr: Yes.

Suppose $x$ and $v$ are orthogonal unit vectors, interpreted as a point of the unit sphere and a tangent vector, let $w^{\perp}$ be an arbitrary vector orthogonal to $x$, and set $w = w^{\perp} - x$, so that $\brak{w, x} = -1$.
Define
$$
X(t) = x\cos t + v\sin t + \tfrac{1}{2}t^{2} w^{\perp},\qquad
\alpha(t) = \frac{X(t)}{|X(t)|}.
$$
Since $x\cos t + v\sin t$ is a unit vector for all $t$, we have
\begin{align*}
  |X(t)|^{2}
  &= 1 + t^{2}\brak{x\cos t + v\sin t, w^{\perp}} + \tfrac{1}{4}t^{4} |w^{\perp}|^{2} \\
  &= 1 + t^{2}\sin t\brak{v, w^{\perp}} + \tfrac{1}{4}t^{4} |w^{\perp}|^{2} \\
  &= 1 + O(t^{3})
\end{align*}
at $t = 0$. Since $(1 + u)^{-1/2} = 1 - \frac{1}{2}u + O(u^{2})$ at $u = 0$, we have $1/|X(t)| = 1 + O(t^{3})$ at $t = 0$, so
$$
\alpha(t) = X(t)[1 + O(t^{3})].
$$
Particularly, $\alpha'(0) = X'(0)$ and $\alpha''(0) = X''(0) = -x + w^{\perp} = w$.
A: The first condition is not necessary.
Let $R(t) \in \mathsf{SO}(3)$ and $\omega(t) = (\omega^1(t), \omega^2(t), \omega^3(t)) \in \mathbb{R}^3.$
Let
$$
S(\omega(t)) := \begin{pmatrix} 0 & \omega^3(t) & -\omega^2(t) \\ -\omega^3(t) & 0 & \omega^1(t) \\ \omega^2(t) & -\omega^1(t) & 0 \end{pmatrix}
$$
Consider the dynamical system,
$$\begin{aligned}
\dot{R}(t) &= S(\omega(t))\, R(t),\\
\dot{\omega}(t) &= u(t)
\end{aligned},$$
where $u(t) \in \mathbb{R}^3$ is any arbitrary signal. Since $R(t)\in\mathsf{SO}(3)$ for all $t,$ we have that,
$$
x(t) := R(t)\,e_1 \in \mathsf{S}^{2},
$$
for all $t.$ This is independent of $u(t).$ It is straightforward to see that  the velocity of $x$ (so-to-speak) is orthogonal to $x(t)$ with the standard inner product. Observe that,
$$\begin{aligned}
\langle x(t), \dot{x}(t) \rangle
&= \langle R(t)\,e_1, S(\omega(t))\,R(t)\,e_1 \rangle\\
&=
(R(t)\,e_1)^\top S(\omega(t)) (R(t)\,e_1)\\
&=
0,
\end{aligned}$$
where the last step follows from the fact that $S(\omega(t))$ is skew-symmetric.
The only constraint on the acceleration profile is that,
$$
\langle \ddot{x}(t), {x}(t) \rangle = -\langle \dot{x}(t), \dot{x}(t) \rangle
$$
or, equivalently,
$$
\frac{\langle \ddot{x}(t), {x}(t) \rangle}{\langle \dot{x}(t), \dot{x}(t) \rangle} = -1.
$$
That is, the "projection" of the acceleration on the position is the speed (of traversal) squared.
So, for example, one could pick $u(t) = (100, 100, 100)^\top$ with the initial condition $\omega(0) = 0$ and $x(0) \in \mathsf{S}^2$ and you will find that the acceleration does not satisfy your necessary condition anywhere yet $\langle x(t), \dot{x}(t) \rangle = 0$ and $x(t) \in \mathsf{S}^2$ for all $t.$
Of course, this differential equation can also be used to describe curves with the initial conditions you specified with an appropriate choice for $\omega(0).$
