This is Example 9.1b in Lee's Introduction to Smooth Manifolds which is about integral curves. He says
Let $W = x\partial/\partial y - y \partial/\partial x$ on $\mathbb{R}^2$. If $\gamma: \mathbb{R} \rightarrow \mathbb{R}^2$ is a smooth curve, written in standard coordinates as $\gamma(t) = (x(t), y(t))$, then the condition $\gamma'(t) = W_{\gamma(t)}$ for $\gamma$ to be an integral curve translates to $$x'(t)\frac{\partial}{\partial x}\Big|_{\gamma(t)} + y'(t)\frac{\partial}{\partial y}\Big|_{\gamma(t)} = x(t)\frac{\partial}{\partial y}\Big|_{\gamma(t)} - y(t)\frac{\partial}{\partial x}\Big|_{\gamma(t)}.$$
How does how obtain the left hand side above for $\gamma'(t)$?
You know from basic geometry of curves that the tangent vector of a curve $\gamma(t)=(x(t),y(t))$ is defined at any point by $\gamma'(t)=(x'(t),y'(t))$. The LHS is just another way of writing this, calling the standar basis vectors of $\mathbb{R}^2$ $\left(\frac{\partial}{\partial x}\right)_{\gamma(t)}$ and $\left(\frac{\partial}{\partial y}\right)_{\gamma(t)}$, respectively (remember that the tangent space changes with the point; in the trivial case of $\mathbb{R}^2$ we usually forget about this, but the author is stressing this point-dependence with the suscript $\gamma(t)$ below the partial derivatives).