I've come accross this physical interpretation for $ [X,Y] $ which I don't understand :
In the limit as $\epsilon$ approaches 0, the result of the above motion approaches the Lie Bracket $[X,Y]$.
Maybe someone can elucidate this for me?
The phrase 'flow $g$ along $Y$ a small distance $\epsilon$' is very nice, but all it means ultimately is just 'take the directional derivative of the function (e.g. surface) $g$ along $Y$'. Thus $X(Y)g$ means 'take the directional derivative of $Yg$ along $X$'. But $Yg$ is itself a directional derivative. Thus, $X(Y)g$ means 'take the directional derivative along $X$ of the directional derivative of $g$ along $Y$'.
Intuitively this is a generalisation of $\frac{\partial^2 g}{\partial x \partial y}$, since in the Lie bracket the two vector fields $X$ and $Y$ do not have to be orthogonal.
The second half of the Lie bracket then subtracts the same derivations in reverse order. If the two derivations commute, the Lie bracket is zero.
The vector flow terminology has definite aesthetic appeal (my original background is in fluid mechanics), but it remains difficult for me to visualise intuitively how the four arcs shrinking to a point as $\epsilon\rightarrow 0$ end up with this double derivation. The slope of a surface $g(x,y)$ in a particular direction at a given point $(x,y)$, on the other hand, is immediately obvious; and how that slope then may change in the direction of the other vector field is also intuitively clear. Once this has been established, the fact that these two double derivations may differ holds no mysteries.
Of course, the picture of the four arcs and whether they close or not remains the best visualisation of the Lie bracket once what it means has been understood.
To apply a vector field $V$ to a function $g$ (at a point $p$), take the directional derivative of $g$ along $V$. This is to say flow $g$ along $V$ some small distance $\epsilon$, take the difference quotient, and let $\epsilon \to 0$.
The Lie bracket $[X,Y]$ is defined as the vector field given by $[X,Y]f = X(Yf) - Y(Xf)$. So, loosely speaking, we are infinitesimally flowing along $Y$, then $X$ and also along $X$, then $Y$, and taking the difference. Since subtracting is adding the opposite, we're flowing infinitesimally along $Y$, $X$, then $-X$, and finally $-Y$.
I'm assuming you want an informal description, not a formal reason to think that $[X,Y]$ is the flow along $X$, then $Y$, then back along $X$, then back along $Y$. If you want the latter, I'll have to rewrite this.
Write $\mathrm{Fl}^X_t(m)$ for the flow of a vector field $X$, that is $\forall m\in M,\forall t\in I_m$ where $I_m$ is the maximal interval on which the flow through $m$ is defined $$\frac{d}{dt}\mathrm{Fl}^X_t(m)=X_{\mathrm{Fl}^X_t(m)}$$ $$\mathrm{Fl}^X_0(m)=m.$$ Fix a point $m$ on your manifold $M$, and consider the curve $$c:(-\epsilon,+\epsilon)\rightarrow M,~t\mapsto c(t)=\mathrm{Fl}^Y_{-t}(\mathrm{Fl}^X_{-t}(\mathrm{Fl}^Y_t(\mathrm{Fl}^X_t(m))))$$ defined on a small neighborhood of $0$. This curve is what you get if you first follow the flow of $X$ for $t$ seconds, then the flow of $Y$ for $t$ seconds, the go back $t$ seconds along the flowlines of $X$ and go back again $t$ seconds along the flow lines of $Y$. This curve is drawn on the manifold, and passes through $m$ at $t=0$. It is also smooth because the flows of $X$ and $Y$ are smooth.
You can ask for the question "what is its derivative?", in particular what is $$\frac{d}{dt}\bigg|_{t=0}c(t)?$$ Since we are differentiating a curve, the result will be a tangent vector at $c(0)=m$. It turns out, but this requires some calculations, that the answer is $$\frac{d}{dt}\bigg|_{t=0}c(t)=[X,Y]_m.$$
