Calculating the differential of the quotient map using curves We can view the projective space $P(\mathbb R^n)$ as the quotient of $S^n/\sim$  where $x \sim y$ if and only if $x = -y$.
The quotient map $F: S^n \to P(\mathbb R^n)$ is the map $x \mapsto [x]$ where $[x] = \{x,-x\}$.
I am trying to calculate the differential of $F$ at a point $x$. I will denote it $d_xF$. First, I tried to do it  for $S^2$. 
Here is what I did in the case $n=2$:
By definition, $d_xF = {d\over dt}(F\circ \gamma)(0)$ where $\gamma : (-1,1) \to S^2$ is a curve such that $\gamma (0) = x$ and ${d\over dt}\gamma (0) = v$ (here $v$ is a tangent vector at $x$).
My first confusion arose from the definition: I cannot calculate $F'$ since $F$ is not a map on $\mathbb R^n$ but a map between arbitrary manifolds. Yet, the definition requires (by applying the chain rule of the derivative) that we calculate $F'$ since $dF = F'\circ \gamma (0) \cdot \gamma '(0)$.

What am I missing? Why do we not need to calculate $F'$?

I cheated around this problem by substituting in the curve and then taking the derivative as follows:
The first curve I used is $\gamma_1: (-1,1) \to \S^2$, $(\cos t, \sin t , 0)$. Then $\gamma_1(0) = (1,0,0)$ and $(F \circ \gamma)' (0) = [(-\sin t, \cos t, 0)]$ at $0$ which is equal to $[(0,1,0)]$.
Next I used $\gamma_2(t) = (0,\cos t , \sin t)$ so that $dF(0,1,0) = [(0,0,1)]$ and 
$\gamma_3(t) = (\sin t, 0 , \cos t)$ so that $dF(0,0,1) = [(1,0,0)]$.
Therefore
$$ d_xF = \left ( \begin{matrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}\right )$$

Is this correct?

And if so, 

How can I generalise to the case of arbitrary $n$?

 A: I think the basic problem you have (in either approach) is that you need a realization of the tangent spaces of $P(\mathbb R^{n+1})$ before you can try to calculate the derivative. You start with $S^n$ as a submanifold of $\mathbb R^{n+1}$ so you view its tangent spaces as linear subspaces of $\mathbb R^{n+1}$. But $P(\mathbb R^{n+1})$ can certainly not be embedded into $\mathbb R^{n+1}$ (and viewing projective spaces as submanifolds is notoriously complicated). So the easier way will be to view it as an abstract manifold. To do this, you can either intorduce local coordinates as suggested in the comment of @John_Ma or view tangent vectors as derivations acting on smooth functions. Finally, you can also use that the map $F$ you define is a local diffeomorphism. This means that you can view te tangent bundle $TP(\mathbb R^{n+1})$ as $TS^n/\sim$ where the equivalence relation is defined by $(x,v)\sim (-x,-v)$. In this picture, $F'$ actually is the natural way to identify $T_xS^n$ with $T_{F(x)}P(\mathbb R^{n+1})$. 
You could also follow the approach you started, extending the quotient projection to an open neighborhood of $S^n$. But in any case, you cannot hope to get any "formula" for $F'$ without specifying an explicit description of the tangent spaces of $P(\mathbb R^{n+1})$. 
A: Let me expand the excellent answer from Andreas Cap a bit more.
I guess the only confusion is what you called your first confusion: You need to understand what the differential of a map $F: M \to N$ between abstract manifolds $M,N$ does. Your $dxF$ is actually a linear map between the tangent spaces $T_x S^2 \to T_{[x]} P(\Bbb R^3)$. So you first need to understand what those tangent spaces are.
There are multiple equivalent ways to define tangent vectors. One is to say a tangent vector $X$ at $p \in M$ is a $\Bbb R$-linear map $X: C^{\infty} (M) \to \Bbb R$ for which $X(fg) = f(p) X(g) + X(f)  g(p)$ holds. Such a map is called a derivation. One can show, that the derivations at $p$ form a vector space: the tangent space at $p$. In this sense the differential $d_p F$ applied to a tangent vector $X$ needs to be a derivation again, so we need to say, what it does to a functions $g \in C^{\infty} (N)$. The definition is: $d_p F (X) (g) = X(g \circ F)$.
Another way to define tangent vectors is to say, that a tangent vector is an equivalence class of smooth curves $\gamma: [- \epsilon, \epsilon] \to M$, where to curves $\gamma, \delta$ are equivalent, iff $(x \circ f)'(0) = (x \circ g)'(0)$, where $(U,x)$ is a chart around $p$. Note, that $x \circ f$ is a curve $\Bbb R \to \Bbb R^{\dim M}$, so you just need to know real differentiation. (Why is this definition well defined, e.g. independet of the choice of (compatible) charts?) Usually one writes than $X = d/dt|_{t=0} {\gamma} (t) \in T_p M$. In this sense, the differential $d_p F$ apllied to $X$ needs to be another equivalence class of curves and this is done as follows: $d_p F (X)$ is the equivalence class of the curve $F \circ \gamma: [-\epsilon, \epsilon] \to N$. (Again: Why is this well defined?) This amounts to what you have written down: $d_p F (d/dt|_{t=0} \gamma (t)) = d/dt|_{t=0} (F \circ \gamma) (t)$.
That these two definitions are in deed equivalent is contained in every differential geometry book. Furthermore given a chart $(U,x)$ around $p$ there is the standard bases $\frac{\partial}{\partial x^i} \big\vert_p$ ($i = 1, \dots , \dim M$), where for $f \in C^{\infty} (M)$ the thangent vector $\frac{\partial}{\partial x^i} \big\vert_p$ sends $f$ to the $i$th partial derivative of $f \circ x^{-1}$ at $x(p)$. So using charts is the only way to write the differential $d_p F$ as a matrix! In your case, you need to specify a chart around $x \in S^2$ (the stereographic projections for example) and a chart around $[x]$ in $P( \Bbb R^3)$ and than you can write $d_x F$ as a matrix with respect to the standard basis of $T_x S^2T$ and $T_{[x]} P( \Bbb R^3)$ in terms of your chosen charts.
