Constructing a smooth vector field on $\mathbb{RP}^2$ whose coordinate representation wrt a chart equals a given smooth vector field on $\mathbb{R}^2$

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Consider $\mathbb{R}P^2$ and the chart $\varphi_3:U_3 \rightarrow \mathbb{R}^2$ defined by $\varphi_3([x_0:x_1:x_2])=(\frac{x_0}{x_2},\frac{x_1}{x_2})$ where $U_3=\{[x_0:x_1:x_2] \in \mathbb{R}P^2:x_2 \neq 0\}$. Show there exists a smooth vector field $Y \in \mathfrak{X}(\mathbb{R}P^2)$ whose coordinate representation with respect to $\varphi_3$ equals $X=x\frac{\partial}{\partial y}-y \frac{\partial}{\partial x} \in \mathfrak{X}(\mathbb{R}^2)$. Here, it's unspecified but I believe $x$ and $y$ are taken to be the standard coordinates on $\mathbb{R}^2$.

I have made the following attempt at this question. Given $p=[x_0:x_1:x_2] \in U_3$, I define $Y_p=d {\varphi_3}^{-1}_{\varphi_3(p)} \circ X_{\varphi_3(p)}$ which is an element of $T_p \mathbb{R}P^2$. I can explicitly compute the image under $Y$ of an element $p$, since $d{\varphi_3}_{\varphi_3(p)} = \begin{bmatrix} \frac{1}{x_2} & 0 \\ 0 & \frac{1}{x_2} \end{bmatrix}$ hence $d{\varphi_3^{-1}}_{\varphi_3(p)} = \begin{bmatrix} x_2 & 0 \\ 0 & x_2 \end{bmatrix}$. Thus \begin{equation*} Y_p=\begin{bmatrix} x_2 & 0 \\ 0 & x_2 \end{bmatrix}\begin{bmatrix} -\frac{x_1}{x_2} \\ \frac{x_0}{x_2} \end{bmatrix}=\begin{bmatrix} -x_1 \\ x_0 \end{bmatrix}=x_0\frac{\partial}{\partial y}|_p -x_1 \frac{\partial}{\partial x}|_p. \end{equation*} Since the components of $Y$ (the coefficients of the tangent space basis vectors) are smooth, I find that $Y$ is smooth. I want to argue that, with respect to $\varphi_3$, $Y$ is equal to $X$ but it's unclear to me how to do this. Since $d{\varphi_3}_{\varphi_3(p)}$ is an isomorphism of vector spaces, and induced by my chart, it intuitively feels like $Y_p$ and $X_{\varphi_3(p)}$ should be equal after taking coordinates. Advice on whether this intuition is correct, and how one would formalize this claim would be welcomed.

In addition, as I have currently defined it $Y:U_3 \rightarrow T \mathbb{R}P^2$. To show $Y \in \mathfrak{X}(\mathbb{R}P^2)$ I need to extend, in a smooth way, the domain of $Y$ to $\mathbb{R}P^2$. Is this indeed the case, and if so, can you suggest/hint how this may be done?