Dtyjlui

Let $f:V_1timesdotstimes V_N to W$ be multilinear. Then $f$ s differentiable and
$$df(a_1,dots,a_n)(h_1+dots+h_n)=f(h_1,a_2,dots,a_n)+f(a_1,h_2,a_3,dots,a_n)+f(a_1,dots,a_{n-1},h_n)tag{1}$$

Problem: Calculate the derivative of the cross product $mathbb R^3times mathbb R^3 to mathbb R^3$.

Solution: Let $f$ be the cross product. It's easy to show that $f$ is multilinear by a direct calculation thus we can use (1). We calculate the derivative at $(u,v)=((u_1,u_2,u_3),(v_1,v_2,v_3))$. We then have
$$df(u,v)(h_1+h_2)=h_1times v + utimes h_2tag{2}$$
Let $e_i, i=1,2,3$ be the unit vectors in $mathbb R^3$. We calculate
$$e_1times v=begin{pmatrix}0\ -v_3 \ v_2end{pmatrix}$$
$$e_2times v=begin{pmatrix}v_3\ 0 \ -v_1end{pmatrix}$$
$$e_3times v=begin{pmatrix}-v_2 \ v_1 \ 0end{pmatrix}$$
For $u$ we use the identity $utimes e_i=-e_itimes u$. So we get the following derivative matrix:
$$df(u,v)=begin{pmatrix}0&v_3&-v_2&0&-u_3&u_2 \ -v_3 & 0 & v_1 & u_3 & 0 & -u_1 \ v_2 & -v_1 & 0 & -u_2 & u_1 & 0end{pmatrix}$$

Question:
1. Apprently, the $+$ means writing the result of $f$ as a column, so it's not the $+$ from e.g. $mathbb R^3$. Should I see that from the definition above without the example? How should I know the dimension of the imge of $df$ without them specifying it somehow?

2. What $+$ is used in $f(a_i)(h_1+h_2)$? How does $h_1$ and $h_2$ and $h_1+h_2$ look like?

For a multilinear mapping, it suffices to consider its Frechet derivative. Let $W$ be an $n$-D vector space, and each $V_i$ be an $m_i$-D vector space with $i=1,2,...,N$. Let $f:V_1times V_2timescdotstimes V_Nto W$ be multilinear. Then $forallleft(v_1,v_2,...,v_Nright)in V_1times V_2timescdotstimes V_N$, the Frechet derivative of $f$ at this location, denoted by $({rm d}f)(v_1,v_2,...,v_N)$, is also a multilinear mapping, i.e.,
$$
({rm d}f)(v_1,v_2,...,v_N):V_1times V_2timescdotstimes V_Nto W.
$$
According to Frechet, it follows that
begin{align}
&({rm d}f)(v_1,v_2,...,v_N)(h_1,h_2,...,h_N)\
&=f(h_1,a_2,...,a_N)\
&+f(a_1,h_2,...,a_N)\
&+cdots\
&+f(a_1,a_2,...,h_N).
end{align}

Recall that, if $g$ is linear, its entry-wise form reads
$$
g_i(v)=sum_ja_{ij}v_j,
$$
and if $g$ is bilinear, its entry-wise form reads
$$
g_i(v_1,v_2)=sum_{j_1,j_2}a_{ij_1j_2}v_{1j_1}v_{2j_2}.
$$
Inductively and formally, the above multilinear $f$ observes the following entry-wise form
$$
f_i(v_1,v_2,...,v_N)=sum_{j_1=1}^{m_1}sum_{j_2=1}^{m_2}cdotssum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}v_{1j_1}v_{2j_2}...v_{Nj_N}
$$
for $i=1,2,...,m$, where each $v_{kj_k}$ denotes the $j_k$-th entry of $v_kin V_k$, while $a_{ij_1j_2...j_N}$'s are the coefficients of $f$.

Thanks to this entry-wise form, we may then write down the entry-wise form of ${rm d}f$ as well, which reads
begin{align}
&({rm d}f)_i(v_1,v_2,...,v_N)(h_1,h_2,...,h_N)\
&=sum_{j_1=1}^{m_1}sum_{j_2=1}^{m_2}cdotssum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}h_{1j_1}v_{2j_2}...v_{Nj_N}\
&+sum_{j_1=1}^{m_1}sum_{j_2=1}^{m_2}cdotssum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}v_{1j_1}h_{2j_2}...v_{Nj_N}\
&+cdots\
&+sum_{j_1=1}^{m_1}sum_{j_2=1}^{m_2}cdotssum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}v_{1j_1}v_{2j_2}...h_{Nj_N}.
end{align}
In other words, as $a_{ij_1j_2...j_N}$'s are known, the entry-wise form of ${rm d}f$ could be expressed straightforwardly as above.

Finally, the "$+$" in OP's original post, i.e., $(h_1+h_2+cdots+h_N)$, is a convention in some context, which is exactly $(h_1,h_2,...,h_N)$ here. When there is free of ambiguity, both expressions can be used as per ones preference.