Given is a scalar function $f$ of 2 vector arguments, $f=f(\vec{x},\vec{y})$. I was unsure if I can say
$$f:\mathbb{R}^{2n} \rightarrow \mathbb{R},$$
but this was the best idea I had and decided to write it here to receive feedback. Under this assumption, it is a Taylor expansion of a function with one vector argument with $2n$ components.
This is an attempt to write the expansion:
Unsure if this expansion is correct... (the expansion is about $\vec{a}=0$, $\vec{b}=0$)
$$f(\vec{x}+\vec{a},\vec{y}+\vec{b})=$$
$$=f(\vec{x},\vec{y})+\left[\frac{\partial f}{\partial x_1}a_1 + \frac{\partial f}{\partial x_2}a_2 + ... +\frac{\partial f}{\partial x_n}a_n + \frac{\partial f}{\partial y_1}b_1+\frac{\partial f}{\partial y_2}b_2+...+\frac{\partial f}{\partial y_n}b_n \right]+...=$$
$$=f(\vec{x},\vec{y})+\left[ (\vec{a}\cdot\nabla_x)f+(\vec{b}\cdot\nabla_y)f \right]+...,$$
where
$$\nabla_x = (\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},...,\frac{\partial}{\partial x_n}),$$
$$\nabla_y = (\frac{\partial}{\partial y_1},\frac{\partial}{\partial y_2},...,\frac{\partial}{\partial y_n}),$$
Is the expansion correct? Pls note this is an expansion of a function of 2 vector arguments; the referenced duplicate question is for a function of 2 scalar arguments - the latter doesn't translate directly to this case, since this case is more complex.
Another worry I had is the $\nabla_y$ I got in the expansion. This 'gradient' is very unusual, since mostly we see $\nabla$ to mean what above is written as $\nabla_x$.
If you define $\vec{z}=(\vec{x},\vec{y})$ and $\vec{c}=(\vec{a},\vec{b})$ then you have \begin{align} f(\vec{x}+\vec{a},\vec{y}+\vec{b})&=f(\vec{z}+\vec{c})\\ &=f(\vec{c})+\langle\nabla f(\vec{c}),~\vec{z}\rangle+\ldots\\ &=f(\vec{a},\vec{b})+\left\langle\begin{pmatrix}\nabla_xf(\vec{a},\vec{b})\\\nabla_yf(\vec{a},\vec{b})\end{pmatrix},~\begin{pmatrix}\vec{x}\\\vec{y}\end{pmatrix}\right\rangle+\ldots\\ &=f(\vec{a},\vec{b})+\langle\nabla_xf(\vec{a},\vec{b}),~\vec{x}\rangle+\langle \nabla_yf(\vec{a},\vec{b}),~\vec{y}\rangle+\ldots\\ &=f(\vec{a},\vec{b})+(\vec{x}\cdot\nabla_x)f(\vec{a},\vec{b})+(\vec{y}\cdot \nabla_y)f(\vec{a},\vec{b})+\ldots \end{align} as an expansion around $(\vec{x},\vec{y})=(0,0)$.