Let $x$ and $y$ be two elements of a Lie group $G$. In chapter 2 of the text "Lie Groups and Lie Algebras I" by A. L. Onishchik, the author states that, if $\overline{x}$ and $\overline{y}$ denote the column vectors of coordinates of $x$ and $y$, respectively, then $\overline{xy} = \overline{x} + \overline{y} + \alpha(\overline{x}, \overline{y}) + \mathcal{O}(\vert\overline{x}\vert^3 + \vert\overline{y}\vert^3)$, where $\alpha$ is a bilinear vector-valued form.
How can I derive this expression? Is it simply the multivariate Taylor series in local coordinates for multiplication function $\mu: G \times G \rightarrow G$ on the product manifold $G \times G$?
Now I think I get it. (I was writing this as a question, but then I think it became an answer to your question. So here it is.)
Updated answer:
They say:
1) Let some coordinate patch near the identity $e \in G$ be given so that $e$ has coordinates zero and so $m : \mathbb{R}^n \to \mathbb{R}^n \to \mathbb{R}^n$ becomes the group multiplication.
Commentary: Multiplication may leave the coordinate chart. Let's just pretend it doesn't, and anyway this can be fixed by shrinking the chart on the domain.
2) In the coordinate patch, we compute the Taylor series expansion $m(x,y) = x + y + a(x,y) + \ldots$, where $a$ is a Bilinear form.
Commentary: I think I understand this. We are computing the Taylor expansion of multiplication around $(0,0)$, so the constant term is zero. Then the linear term comes from the formula $(m(\alpha,\beta))'(0) = \alpha'(0) + \beta'(0)$, which I think is telling us that the differential of multiplication in a Lie group is just addition. So the linear term in the Taylor series expanion is a block matrix $[I,I]$. I think so anyway.
3) Next use the relation $[x,y]yx = xy$ to get that the second order term of $[x,y]$ is $a(x,y) - a(y,x)$.
Commentary: I get that $[x,y] = a(x,y) - a(y,x) - a([x,y],yx) + \ldots$, where the dots are degree $\geq 3$ terms. I don't clearly understand why the $a([x,y],yx)$ term is thought of as being higher order, though I guess it makes sense. It's not clear because $yx$ and $[x,y]$ are multiplied in the Lie group. I guess that one can write out everything to see $a([x,y],yx) = a(x + y - x - y + $g_2(x,y)$, y + x) = a(g_2(x,y), y + x)$, where $g_2$ collects the degree $\geq 2$ terms of $[x,y]$, so that actually $a([x,y],yx)$ lives in the degree $\geq 3$ piece of the Taylor expansion. Okay, that shows their claim.
Old answer: Okay I am starting to get it:
(Edit: No, I think I am still confused.)
As you said, this is the Taylor series expansion of the map $m : G \times G \to G$.
We rewrite this in some local chart as $m : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$, and note the following:
1) $m(0,0) = 0$. ($e* e = e$).
2) $dm(0,0)(x,y) = x + y$, this follows from $(1)$ $(\alpha \beta)'(0) = \alpha'(0) + \beta'(0)$, for $\alpha$ and $\beta$ one parameter subgroups. In particular, $dm(0,0)(x,y)$ is the tangent vector of $\alpha \beta$ at time zero, for $\alpha$ a curve passing through $0$ with velocity $x$ and $\beta$ having velocity $y$.
(See here for an easy proof of (1): https://mathoverflow.net/questions/62898/lie-group-operation-and-tangent-vectors )
Then the Taylor series expansion says exactly that:
$m(x,y) = m(0,0) + dm_{(0,0)}(x,y) + \alpha(x,y) + \text{higher order}$, which is their assertion, given $1)$ and $2)$ above.
Except (and maybe this is what Qiachu was alluding to), there seems to be this implicit identification with the point $x$ with the tangent vector... which is resolved if we let the neighborhood get smaller and smaller, I guess.
I was confused by the equality $dm_{(0,0)}(x,y) = \Sigma \frac{\partial m}{\partial z_i}(0) z_i$, with $(x,y) = (z_i)$, which is the way that I am used to Taylor series being expressed. But these two things are equal, the former being the (vector valued) directional derivative in the direction $(x,y)$ and the latter, well, the same thing.
(Obviously I just don't understand Taylor series!)
Edit:
I also don't understand their derivation of the Taylor series for the commutator, from the equation $[x,y] yx = xy$. One can get again from (1) that the first two terms of the Taylor series for $[x,y]$ are zero. For the degree 2 stuff... maybe there is a similar formula to 1?