I have the following nonlinear system:
$$\begin{pmatrix}\dot{y}_1\\\dot{y}_2\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$
Which I set up to $F=\dot{y}$
Giving the jacobian of transformation being:
$$\dot{F}= \begin{pmatrix}2&0\\2y_1&0\end{pmatrix}=\begin{pmatrix}0&0\\2&0\end{pmatrix}\begin{pmatrix}F_1\\F_2\end{pmatrix}+\begin{pmatrix}2\\0\end{pmatrix}$$
Which gives me the eigenvalues $\lambda=0^2=0,0$ meaning a degenerate node since we have only the eigenvector:
$$F^{(1)}=\begin{pmatrix}0\\1\end{pmatrix}$$
So we have a straight line going up and down from $(2,0)$ I would have thought, but instead it is still coming from $x=0$ as shown below(from here):
It would be better to attack the problem directly as Amzoti has said:
$$\begin{pmatrix}\dot{y}_1\\\dot{y}_2\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$
We want the critical points of this system, which we can find by taking:
$$\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$
$(0,\alpha)$ are the spots of the critical points, thus there are infinite critical points at $x=0$
Furthermore, you have made a mistake in your Jacobian.
Since we are taking $F=\dot{y}$, we need to be very clear that this means:
$$\begin{pmatrix}F_1\\F_2\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$ And hence: $$\dot{F}=\begin{pmatrix}\dot{F}_1\\\dot{F}_2\end{pmatrix}=\begin{pmatrix}2&0\\2y_1&0\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}F_1\\F_2\end{pmatrix}+\begin{pmatrix}2\\0\end{pmatrix}$$