For two $\mathcal{C}^1$ functions $\vec g: \mathbb{R}^n \mapsto \mathbb{R}^m$ and $\vec f: \mathbb{R}^m \mapsto \mathbb{R}^k$ does the equality hold
$$\mathbf{J}_{\vec f} \mathbf{J}_{\vec g} \approx \vec f \circ \vec g$$
where $\approx$ denotes locally linear approximation?
If $\textbf{J}_f$ is evaluated at $g(x)$, then yes. Letting $\textbf{J}_f(x)$ denote the Jacobian evaluated at $x\in\mathbb{R}^m$, the correct approximation is $(f\circ g)(x)\approx \textbf{J}_f(g(x))\cdot\textbf{J}_g(x)$.
For a differentiable function $h:\mathbb{R}^n\to\mathbb{R}^m$, the best linear approximation of $h$ at a point $a\in\mathbb{R}^n$ is defined rigorously with the total derivative, which is a linear map $(Dh)(a):\mathbb{R}^n\to\mathbb{R}^m$ that best resembles the behavior of $h$ near $a$. If all the first-order partial derivatives of $h$ are continuous at $a$, the map $(Dh)(a)$ is just the Jacobian at $a$: $(Dh)(a)=\textbf{J}_h(a)$.
The total derivative has a chain rule; assuming any necessary domain and codomain restrictions, we have $D(f\circ g)(x)=(Df)(g(x))\circ(Dg)(x)$. When the linear maps are given by matrices, the composition becomes matrix multiplication, explaining where the approximation $(f\circ g)(x)\approx \textbf{J}_f(g(x))\cdot\textbf{J}_g(x)$ comes from.