Question: Suppose that $a,b,c,d$ are vectors in $\mathbb{R}^2$, so $a\_{\alpha}$ where $\alpha = 1,2$. We define a scalar product as $a \cdot b := a\_1 b\_2 - a\_2 b\_1$. Suppose also that we have a real-valued function $f(a,b,c)$ which is a function of some $3$ vectors.

What does the Taylor expansion of $f$ look like to first order? For example, I want to Taylor expand $f(a+b,c,d)$ around $(a, c, d)$, so $$f (a+b, c, d) = f(a,c,d) + \ldots $$ but I don't know how to write the remaining terms. Do I get a scalar product of $b$ with some term?

Answer: 
$\def\va{{\bf a}}
\def\vb{{\bf b}}
\def\vc{{\bf c}} 
\def\vd{{\bf d}}
\def\e{\varepsilon}
\def\np{\odot}$We have
\begin{align\*}
f(\va+\vb,\vc,\vd) &= f(\va,\vc,\vd) + \sum\_i b\_i\left.\frac{\partial f(\va+\vb,\vc,\vd)}{\partial b\_i}\right|\_{b\_i=0} + \ldots \\
&= f(\va,\vc,\vd) + \sum\_i b\_i \frac{\partial f(\va,\vc,\vd)}{\partial a\_i} + \ldots \\
&= f(\va,\vc,\vd) + (\vb,\nabla\_\va f(\va,\vc,\vd)) + \ldots, 
\end{align\*}
where $(,)$ is the usual inner product and where
$\nabla\_\va f = \langle \partial f/\partial a\_1,\partial f/\partial a\_2\rangle$ is the gradient.
But
$(\langle a\_1,a\_2\rangle,\langle b\_1,b\_2\rangle)
= \langle a\_1,a\_2\rangle\np\langle -b\_2,b\_1\rangle$.
(To avoid confusion with the dot product we will write $\va\np\vb=a\_1b\_2-a\_2b\_1$.)
Thus,
\begin{align\*}
f(\va+\vb,\vc,\vd) &= f(\va,\vc,\vd) + \vb\np \widetilde\nabla\_\va f(\va,\vc,\vd) + \ldots, 
\end{align\*}
where
$\widetilde\nabla\_\va f = \langle -\partial f/\partial a\_2,\partial f/\partial a\_1\rangle$.

**Addendum**

As is discussed in detail in the other answers, the product $\va\np\vb$ is only a "scalar product" in the sense that it is a product of vectors returning a real value.
If have only a partial answer, hope this helps:

The difficulty here is definitely your "scalar product". I set this in quotation marks because it is not a scalar product (not even a semi-definite one) because it is skew-symmetric: $a\cdot b = -b \cdot a$  

It is merely a skew-symmetric bilinear form, which does not define a metric on your space. Now, I do not know much about exotic topologies, but I am pretty sure you need at least a sense of distance (i.e., a metric) to define derivatives on any space.  

This is not given by your scalar product and therefore the Taylor expansion is not defined.

Is it possible that your "scalar product" is independent from the geometry of the space? I.e., it is just a skew-symmetric bilinear form defined in $\mathbb{R}^2$?  

Then, the Taylor expansion would be the usual one with the usual scalar product $\langle a, b \rangle = a\_1 b\_1 + a\_2 b\_2$.

Let me know if I am wrong with anything!