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: 
There is something to confront here in your question, which has been pointed out already. The first thing is to see that your "scalar product" is not a scalar product: it's a skew symmetric bilinear form. More specifically, notice that $a \cdot a = 0 \hspace{4pt} \forall a$. What you've written is actually the magnitude of the vector-cross product of $a$ and $b$ (More precisely, given a linear map $T: \mathbb{R}^2 \to \mathbb{R}^3 $, $a \cdot b = \vert Ta \times Tb \vert$). This is not a scalar product.

We can relate your skew symmetric form to the usual Euclidean inner product $( \cdot, \cdot)\_E $ by the formula
$$
a \cdot b = (a, Jb)\_E
$$

Where $J$ is a matrix,
$$
J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
$$

So that's what we're dealing with. Now let's look at a Taylor expansion of your function. Any sensible Taylor expansion should look like
$$
F(a + tb) = F(a) + (b \cdot DF(a)) t + \mathcal{O}(t^2)
$$
Where $DF(a)$ is some suitable notion of derivative at $a$.
From this we can see that
$$
b \cdot DF(a) = \frac{d}{dt} F(a + tb) \vert\_{t=0}
$$
I think that formula should get at the crux of your question. However, there's more to notice here. We can write out the right side as the usual inner product with the gradient $ \nabla F $:
$$
\frac{d}{dt} F(a + tb) \vert\_{t=0} = (b, \nabla F(a))\_E
$$
and using our relation of the skew symmetric form to the euclidean inner product, we have
$$
b \cdot DF(a) = (b, J DF(a))\_E
$$
These two are equal, and if they are equal for all $b$, this implies
$$
J DF(a) = \nabla F(a)
$$
$J$ is invertible, so we see that $ DF(a) = J^{-1} \nabla F(a)$. Summarizing,
$$
F(a+b) = F(a) + b \cdot (J^{-1} \nabla F(a)) + \mathcal{O}(t^2)
$$
$\def\va{{\bf a}}
\def\vb{{\bf b}}
\def\vc{{\bf c}} 
\def\vd{{\bf d}}
\def\e{\varepsilon}
\def\np{\odot}$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.