Tangent vector to a curve $\gamma:(-\pi,\pi)\to \mathbb P^2(\mathbb R)$ Let $\gamma: (-\pi,\pi)\to \mathbb P^2(\mathbb R)$ such that $\gamma(t)=(1:\sin t:\cos t)$. I have to find the tangent vector $v\in T_{p_o} \mathbb P^2(\mathbb R)$ to $\Gamma=\gamma((-\pi,\pi))$ with $p_0=\gamma(0).$
Then, given $F:\mathbb P^2(\mathbb R)\to \mathbb P^5(\mathbb R)$ s.t. $F(x^0:x^1:x^2)=((x^0)^2:x^0x^1:x^0x^2:(x^1)^2:x^1x^2:(x^2)^2)$, I have to show that $F_{*,p_0}(v)$ is tangent to $F(\Gamma)$ in $F(p_0)$.
To cover $\mathbb P^2(\mathbb R)$ I chose the standard atlas $\{(U_i,\varphi_i)\}_{i=0}^2$, where $U_i=\{(x^0:x^1:x^2)\in \mathbb P^2(\mathbb R)|x^i\ne 0\}$ and $\varphi_i:\mathbb P^2(\mathbb R)\to\mathbb R^2$ such that $\varphi_i((x^0:x^1:x^2))=(\dfrac{x^j}{x^i},\dfrac{x^k}{x^i})$.
The point $\gamma(0)$ is $(1:0:1)\in U_0\cap U_2$. In order to compute the tangent vector I have to calculate
$$\dot\gamma(0)=\underbrace{\dfrac{d}{dt}\bigg|_{t=0}\gamma^i}_{=a_i}\dfrac{\partial}{\partial y^i}\bigg|_{\gamma(0)}$$
$\gamma^i$ is the composition of $y^i\circ\gamma$, where $y^i$ is the coordinate function.
For example if we take the chart $\varphi_0$, its coordinate functions should be $y^i=\pi^i\circ\varphi_0$ but $\gamma(t)$ has three components, while the coordinate functions are two...
If I have to consider the omitted coordinates, then I'd have
$$x^0\mapsto \hat {\dfrac{x^0}{x^0}}=y^0$$
$$x^1\mapsto \dfrac{x^1}{x^0}=y^1$$
$$x^2\mapsto  \dfrac{x^2}{x^0}=y^2$$
For the second part I did this: I called $\hat F=\psi_0\circ F\circ \varphi_0^{-1}$,where $\psi_0$ and $\varphi_0$ are the charts with domain $V_0$ and $U_0$. Then I called $q_0=\hat F(p_0)$ and I computed
$$\hat F_{*,q_0}(v_0)=\hat F_{*,q_0}(\dfrac{\partial}{\partial x}\bigg|_{p_0})=\langle\dfrac{\partial}{\partial x}\bigg|_{p_0}\hat F^i, \dfrac{\partial}{\partial x^i}\bigg|_{q_0}\rangle$$
 A: We usually speak of the tangent vector to curve $\gamma$ instead of the tangent vector to the image of curve $\Gamma$. For example, fix $v\in \mathbb R^n$ and consider two curves $\gamma_i\colon \mathbb R\to \mathbb R^n$:
$$\gamma_1(t) = vt, \quad \gamma_2(t) = 2vt.$$
Although the images of both curves are the same (it's just a line), $\gamma_1'(0)=v$ and $\gamma'_2(0)=2v$. Hence, the image does not suffice to speak of the tangent vector and we need the whole curve, which gives the information about parametrisation.
Now, if $\gamma\colon (a, b)\to M$ is a curve and $f\colon M\to N$ is a smooth function, we can consider a curve $f\circ \gamma \colon (a, b)\to N$. It holds in general that $f_* \gamma'(t) = (f\circ \gamma)'(t)$. It can be treated either as an easy theorem (if you define vectors as derivations), or even as a part of the definition of $f_*$ (if you define vectors as equivalence classes of curves).
Let's go back to the curve $\gamma\colon (-\pi, \pi) \to \mathbb P^2(\mathbb R)$ with $\gamma(t)=(1 : \sin t : \cos t)$. Indeed, to calculate $\gamma'(0)$ it's the easiest to use one of the standard charts you mentioned:
$$U = \{ (x^0 : x^1 : x^2) \mid x^0\neq 0 \},\quad \phi(x^0 : x^1 : x^2) = \frac{(x^1, x^2)}{x^0},$$
i.e. the coordinate functions $y^i=\pi^i \circ \phi$  are $y^i(x^0: x^1:x^2)=x^i/x^0$ for $i=1, 2$.
Now the function $\phi \circ \gamma: (-\pi, \pi)\to \mathbb R^2$ is given by $t\mapsto (\sin t, \cos t)$. We need $(\phi \circ \gamma)'(0)$, which is an easy exercise in calculus.
