Let $X=\sum_ja_j(x)\frac{\partial }{\partial x_j}$ be a $BV$ vector field in an open subset of $\mathbb R^n.$ Alberti's theorem says that $$ DX_s=(\frac{\partial a_j }{\partial x_k})_{1\le j,k\le n}=S \otimes \eta, \quad \text{$S(x)$ tangent vector at $x$, $\eta(x)$ cotangent vector at $x$} $$ i.e for $T$ tangent at $x$, $ (DX)_s(x) T= \langle \eta(x), T\rangle S(x). $ Intuitively, it means essentially that, near each point, you can find a coordinate system such that each coefficient $a_j$ of your vector field is in fact a function $$ a_j(x', x_n) \text{ such that}\ \frac{\partial a_j}{\partial x'} \in L^1,\quad \frac{\partial a_j}{\partial x_n} \text{is a Radon measure}, $$ where $x'$ is $n-1$ dimensional and $x_n$ has just one dimension.
Let me add a couple of comments. You have a canonical decomposition of $DX$. What matters is the singular part. When I write as above $DX_s=S\otimes \eta$, it means that for a tangent vector $T$, $$\underbrace{DX_s}_{\text{matrix}}T=\langle\eta,T\rangle \underbrace{S}_{\text{vector}}.$$ You may decide that $\eta=e_n^*$ so that $DX_sT=T_n S$ and if you know that the divergence of $X$ is absolutely continuous, you have $S_n=0$, so that you may assume that $S=e_1$. As a result, $$ DX_s e_n=e_1,\quad DX_s e_k=0 \text{ for $1\le k\le n-1$}. $$ Since $DX=\bigl(\frac{\partial a_j}{\partial x_k}\bigr)$, $j$ line-index, $k$ column-index, you get that only the last column $$ \frac{\partial a_j}{\partial x_n}\quad \text{can be a Radon measure},\quad \frac{\partial a_j}{\partial x_k}, k\le n-1\quad \text{(any $j$) is $L^1$} $$ With our assumption on the divergence and our choice $S=e_1$, we have only $$ \frac{\partial a_1}{\partial x_n} \text{singular, other entries $L^1$.} $$ Of course, you have to keep in mind that $S,\eta$ depend on the point $x$, but through an approximation procedure, you may indeed assume that singularity occurs by differentiating only in one direction (and with null divergence (or ac) is concerning only one off-diagonal coefficient).