EDIT: I have posted the solution as an answer. The things I have written below in the question have errors.
I am wondering if I have approached this question correctly from Lee's Smooth Manifolds. Specifically, to show a vector field exists, is all I would need to show that the pushforwards are smooth in each chart and for example don't include a term of 1/x?
(Exercise 8-12) Let $F: \mathbb{R}^2 \rightarrow \mathbb{R} \mathbb{P}^2$ be the
smooth map $F(x, y)=[x, y, 1]$, and let $X \in$
$\mathfrak{X}\left(\mathbb{R}^2\right)$ be defined by $X=x \partial /
\partial y-y \partial / \partial x$. Prove that there is a vector
field $Y \in \mathfrak{X}\left(\mathbb{R P}^2\right)$ that is
$F$-related to $X$, and compute its coordinate representation in terms
of each of the charts defined in Example 1.5.
There are three charts on $\mathbb{R}\mathbb{P}^2$. They are
$$\varphi_1([x,y,z])=\left(\frac{x}{z},\frac{y}{z}\right)$$
$$\varphi_2([x,y,z])=\left(\frac{x}{y},\frac{z}{y}\right)$$
$$\varphi_3([x,y,z])=\left(\frac{y}{x},\frac{z}{x}\right).$$
We compute $F_*X$ in the first chart. Let $q=[x,y,1]$. Then
$$
F_*X
=d_{F^{-1}(q)}F(X_{F^{-1}(q)})
=x\left.\frac{\partial}{\partial y}\right|_{(x,y)}-\left.y\frac{\partial}{\partial x}\right|_{(x,y)}
$$
Now the transition map between $\varphi_1$ coordinates and $\varphi_2$ coordinates is
$$(x,y) \mapsto\left(\frac{x}{y}, \frac{1}{y}\right)=(\tilde x, \tilde y).$$
Now we convert to $(\tilde x, \tilde y)$ coordinates.
$$
\begin{aligned}
\left.\frac{\partial}{\partial x}\right|_{(x, y)} & =\left.\frac{\partial \tilde{x}}{\partial x} \frac{\partial}{\partial \tilde{x}}\right|_{\left(\frac{x}{y}, \frac{1}{y}\right)}+\left.\frac{\partial \tilde{y}}{\partial x} \frac{\partial}{\partial \tilde{y}}\right|_{\left(\frac{x}{y}, \frac{1}{y}\right)} \\
& =\left.\frac{1}{y} \frac{\partial}{\partial \tilde{x}}\right|_{\left(\frac{x}{y}, \frac{1}{y}\right)} \\
& =\left.\tilde{y} \frac{\partial}{\partial \tilde{x}}\right|_{(\tilde{x}, \tilde{y})}
\end{aligned}
$$
and
$$
\begin{aligned}
\left.\frac{\partial}{\partial y}\right|_{(x, y)} & =\left.\frac{\partial \tilde{x}}{\partial y} \frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x}, \tilde{y})}+\left.\frac{\partial \tilde{y}}{\partial y} \frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x}, \tilde{y})} \\
& =-\left.\frac{x}{y} \frac{1}{y} \frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x}, \tilde{y})}+-\left.\frac{1}{y^2} \frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x}, \tilde{y})} \\
& =-\left.\tilde{x} \tilde{y} \frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x}, \tilde{y})}-\left.\tilde{y} \frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x}, \tilde{y})}.
\end{aligned}
$$
So in the second chart, we have the transition map from the first chart as
$$
\begin{aligned}
& x=\tilde{x} y=\frac{\tilde{x}}{\tilde{y}} \quad \text { and } y=\frac{1}{\tilde{y}} \\
& \text { so } \\
& F_* X=-\frac{\tilde{x}}{\tilde{y}}\left(\left.\tilde{y} \frac{\partial}{\partial \tilde{x}}\right|_{(\tilde{x}, \tilde{y})}\right)-\frac{1}{\tilde{y}}\left(-\left.\tilde{x} \tilde{y} \frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x}, \tilde{y})}-\left.\tilde{y} \frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x} \tilde{y})}\right) \\
&=-\left.\tilde{x} \frac{\partial}{\partial \tilde{x}}\right|_{(\tilde{x}, \tilde{y})}+\left.\tilde{x} \frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x}, \tilde{y})}+\left.\frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x}, \tilde{y})} \\
&=-\left.\tilde{x} \frac{\partial}{\partial \tilde{x}}\right|_{(\tilde{x}, \tilde{y})}+\left.(\tilde{x}+1) \frac{\partial}{\partial \tilde{y}}\right|_{(\tilde{x}, \tilde{y})}.
\end{aligned}
$$
For the third chart, we have the transition map from the first chart in standard coordiantes
$$
\begin{aligned}
& (x, y) \mapsto\left(\frac{y}{x}, \frac{1}{x}\right)=(\hat{x}, \hat{y}) \\
\end{aligned}
$$
Then $\hat{x}=\frac{y}{x} \quad \hat{y}=\frac{1}{x}$
$$
\begin{aligned}
\left.\frac{\partial}{\partial x}\right|_{(x, y)} & =\left.\frac{\partial \hat{x}}{\partial x} \frac{\partial}{\partial \hat{x}}\right|_{(\hat{x}, \hat{y})}+\left.\frac{\partial \hat{y}}{\partial x} \frac{\partial}{\partial \hat{y}}\right|_{(\hat{x}, \hat{y})} \\
& =-\left.\frac{y}{x^2} \frac{\partial}{\partial \hat{x}}\right|_{(\hat{x}, \hat{y})}-\left.\frac{1}{x^2} \frac{\partial}{\partial \hat{y}}\right|_{(\hat{x}, \hat{y})} \\
&=-\left.\hat{x} \hat{y} \frac{\partial}{\partial \hat{x}}\right|_{(\hat x, \hat y)}-\left.\hat{y} \frac{\partial}{\partial \hat{y}}\right|_{(\hat{x}, \hat{y})}\\
\end{aligned}
$$
and
$$
\begin{aligned}
\left.\frac{\partial}{\partial y}\right|_{(x, y)} & =\left.\frac{\partial \hat{x}}{\partial y} \frac{\partial}{\partial \hat{x}}\right|_{(\hat{x}, \hat{y})}+\left.\frac{\partial \hat{y}}{\partial y} \frac{\partial}{\partial \hat{y}}\right|_{(\hat x, \hat y)} \\
& =\left.\frac{1}{x} \frac{\partial}{\partial \hat{x}}\right|_{(\hat{x}, \hat{y})} \\
& =\left.\hat{y} \frac{\partial}{\partial \hat{x}}\right|_{(\hat{x}, \hat{y})}.
\end{aligned}
$$
Subsituting these in, we have
$$
\begin{aligned}
F_* X & =\left.x \frac{\partial}{\partial y}\right|_{(x, y)}-\left.y \frac{\partial}{\partial y}\right|_{(x, y)} \\
& =x\left[-\left.\hat{x} \hat{y} \frac{\partial}{\partial \hat{x}}\right|_{(\hat{x}, \hat{y})}-\left.\hat{y} \frac{\partial}{\partial \hat{y}}\right|_{(\hat{x}, \hat{y})}\right]-y\left[\left.\hat{y} \frac{\partial}{\partial \hat{x}}\right|_{(\hat{x}, \hat{y})}\right]
\end{aligned}
$$
and $x=\frac{1}{\hat{y}}, \quad y=\frac{\hat{x}}{\hat{y}}$ so
$$
\begin{aligned}
F_* X & =-\left.\frac{\hat{x} \hat{y}}{\hat{y}} \frac{\partial}{\partial \hat{x}}\right|_{\hat{(x, y)}}=\left.\frac{\hat{y}}{\hat{y}} \frac{\partial}{\partial \hat{y}}\right|_{(\hat{x}, \hat{y})}-\left.\frac{\hat{x}}{\hat{y}} \hat{y} \frac{\partial}{\partial \hat{x}}\right|_{(\hat{x}, \hat{y})} \\
& =-\left.2 \hat{x} \frac{\partial}{\partial \hat{x}}\right|_{(\hat{x}, \hat{y})}-\left.\frac{\partial}{\partial \hat{y}}\right|_{(\hat{x}, \hat{y})}.
\end{aligned}
$$
Since the push-forward exists in every chart, there is an $F$-related vector field.
Best Answer
It looks like this exercise wants us to show that for this $F$ whose range does not include the point $[1,1,0]$ at infinity of $\mathbb {R\,P}^2$ there is a very natural chart $\varphi_1$ which does not lead to any singularities and works for the whole domain and range of $F\,.$
When we use the other charts $\varphi_2,\varphi_3$ we cannot get away without terms $1/x^2$ or $1/y^2\,.$ Also: even when we patch together both, the union of their domains does not contain the point $[0,0,1]$ and the union of their coordinate ranges does not contain the origin $(0,0)$ of $\mathbb R^2\,.$
I redo the calculations in what I believe is a simpler notation (did not check all calculations of yours).
The map $F:\mathbb R^2\to\mathbb {P\,R}^2,(x,y)\mapsto F(x,y)=[x,y,1]\,$ has in the chart $$\tag1 \varphi([x,y,z])=\left(\frac xz,\frac yz\right) $$ the differential (Jacobi matrix) $$\tag2 d(\varphi\circ F)_p=\pmatrix{1&0\\0&1}\,,\quad p=(x,y)\,, $$ because $(\varphi\circ F)(x,y)=(x,y)\,.$
In the chart $\varphi\,,$ the pushforward of $X=x\,\partial_y-y\,\partial_x$ (which is in $T\mathbb R^2$) by $F$ is \begin{align} (F_*X)_{(\varphi\,\circ\, F)(p)}&=d(\varphi\circ F)_p(X)=\underbrace{\pmatrix{1&0\\0&1}\pmatrix{-y\\x}}_{\textstyle {-y\choose x}}\cdot\pmatrix{\partial_x\\\partial_y}_{(\varphi\,\circ\, F)(p)}\\[2mm]\tag3 &=(x\,\partial_y-y\,\partial_x)_{(x,y)} =x\,\partial_y-y\,\partial_x\,. \end{align} I emphasize that this is in $T_{F(p)}\mathbb{P\,R}^2$ and expressed in the first chart $\varphi\,.$
For the second chart $$\tag4 \varphi([x,y,z])=\left(\frac{x}{y},\frac{z}{y}\right) $$ I use the same symbol $\varphi$ to simplify notation. Then, cutting short, \begin{align}\tag5 (\varphi\circ F)(x,y)&=\left(\frac xy,\frac 1y\right)\,,& d(\varphi\circ F)_p&=\pmatrix{\frac 1y&-\frac x{y^2}\\0&-\frac1{y^2}}=\frac{-1}{y^2}\pmatrix{-y&x\\0&1}\,,\end{align}\begin{align}\tag6 (F_*X)_{(\varphi\,\circ\,F)(p)}&=-\tfrac{x^2+y^2}{y^2}\partial_x-\tfrac x{y^2}\,\partial_y\,. \end{align} Likewise, for the third chart \begin{align}\tag7 (\varphi\circ F)(x,y)&=\left(\frac yx,\frac 1x\right)\,,& d(\varphi\circ F)_p&=\pmatrix{\frac{-y}{x^2}&\frac 1x\\-\frac1{x^2}&0}=\frac{-1}{x^2}\pmatrix{y&-x\\1&0}\,,\end{align}\begin{align}\tag8 (F_*X)_{(\varphi\,\circ\,F)(p)}&=\tfrac{x^2+y^2}{x^2}\partial_x+\tfrac y{x^2}\,\partial_y\,. \end{align}