I would to like to calculate the $\nabla f$, where $f$ is a restriction of the function $F: \mathbb{R}^{n+1}\to \mathbb{R} $ to $\mathbb{S^{n}}$.
I know that $\nabla f = \sum_{i,j=1}^{n} g^{ij}\frac{\partial f }{\partial x_{i}} \frac{\partial }{\partial x_{j}}$, where $g^{ij}$ is a inverse matrix of metric coefficients given by stereo graphic projection. By formation law of the function
$\nabla f = \sum_{i,j=1}^{n} g^{ij}\frac{\partial f }{\partial x_{i}} \frac{\partial }{\partial x_{j}}= \sum_{j=1}^{n} g^{1j}\frac{\partial }{\partial x_{j}}$. However, it is a hard to finish. There exists a short hut for calculating the gradient of function?
Let $M$ be a submanifold of $\mathbb R^n$ and $f : \mathbb R^n \to \mathbb R$ is a smooth function. Then
$$\tag{1}\nabla (f|_M) = (\nabla f)^\top,$$
where $\top$ is the projection onto the tangent space of $M$. To see this, note that $\nabla (f|_M)$ is the unique vector fields on $M$ so that
$$ \langle \nabla f|_M , v\rangle = \nabla_v (f|_M), \ \forall v\in TM,$$
where $\nabla_v$ denotes directional derivatives. To check $(1)$, let $x\in M$ and $v\in T_xM$. Then there is a curve $\gamma :(-\epsilon, \epsilon) \to M$ so that $\gamma(0) = x$ and $\gamma'(0) = v$. Thus
$$\nabla_v (f|_M) =\frac{d}{dt} f(\gamma(t))\bigg|_{t=0}.$$
But the right hand side is also (think of $\gamma $ mapping to $\mathbb R^n$)
$$\langle \nabla f, v\rangle. $$
Thus
$$\langle \nabla (f|_M), v\rangle = \nabla_v f = \langle \nabla f, v\rangle = \langle (\nabla f)^\top, v\rangle \ \ \forall v \Rightarrow \nabla (f|_M) = (\nabla f)^\top.$$