If I understood your question correctly, it may help clarifying what is done right in the first step.
Note that the derivative of $z$ w.r.t. to $y$ is a function, i.e.
$$\frac{\partial z}{\partial y}(u, v)$$
makes sense and we can rename it to, say, $f(u,v) = \frac{\partial z}{\partial y}(u,v) = x^2 \frac{\partial z}{\partial u} + 2\frac{\partial z}{\partial v}$.
Now let me derive $f$ w.r.t. to $x$:
$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y} \right) = \frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} + 2\frac{\partial z}{\partial v} \right)$$
Now what was done was, first distribute the derivative w.r.t. $x$:
$$\frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} + 2\frac{\partial z}{\partial v} \right) = \frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} \right) + \frac{\partial}{\partial x}\left(2\frac{\partial z}{\partial v} \right)$$
Now "taking $x$ out" is really the product rule $(ab)' = a'b + ab'$:
$$\frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} \right) = \frac{\partial}{\partial x}(x^2)\times \frac{\partial z}{\partial u} + x^2 \times \frac{\partial}{\partial x}\frac{\partial z}{\partial u} = 2x\frac{\partial z}{\partial u} + x^2 \frac{\partial}{\partial x}\frac{\partial z}{\partial u}$$
From there we just use the chain rule to find $ \frac{\partial}{\partial x}\frac{\partial z}{\partial u}$.
Is it clearer now?