Partial derivative on vector elements with the vector itself

$\tfrac{\partial ~~~}{\partial\vec y}$ is itself a gradient operator, $\nabla_{y_1,y_2}$, and the gradient of a scalar with respect to a 2D vector is a $1\times 2$ matrix (a row vector). So just nest the operations.

$$\begin{align}\nabla_{x,\vec y}~\vec F(x,\vec y)&=\left(\begin{bmatrix}\tfrac{\partial ~~}{\partial x}\\\nabla_{\vec y}\end{bmatrix}\begin{bmatrix}F_1(x,\vec y)&F_2(x,\vec y)\end{bmatrix}\right)^{\top}\\&=\begin{bmatrix}\tfrac{\partial~~}{\partial x} F_1(x,y)&\nabla_{y_1,y_2}F_1(x,\vec y)\\\tfrac{\partial~~}{\partial x} F_2(x,y)&\nabla_{y_1,y_2}F_2(x,\vec y)\end{bmatrix}\\[1ex]&=\begin{bmatrix}\tfrac{\partial~~}{\partial x} F_1(x,y)&\begin{bmatrix}\tfrac{\partial~~~}{\partial y_1}F_1(x,\vec y)&\tfrac{\partial~~~}{\partial y_2}F_1(x,\vec y)\end{bmatrix}\\\tfrac{\partial~~}{\partial x} F_2(x,y)&\begin{bmatrix}\tfrac{\partial~~~}{\partial y_1}F_2(x,\vec y)&\tfrac{\partial~~~}{\partial y_2}F_2(x,\vec y)\end{bmatrix}\end{bmatrix}\end{align}$$

This nesting might be ignored by some. If we define the composition: $\vec G(x,y_1,y_2)=\vec F(x,\vec y)$ , and handwave the gradient operators as being equivalent, then we would claim $\nabla_{x,\vec y}\vec F(x,\vec y)=\nabla_{x,y_1,y_2}\vec G(x,y_1,y_2)$, which is a $2{\times}3$ matrix.

Either way, you can evaluate the six members.