Multiplication formula for Lie derivative 
Let $U\in\mathbb{R}^n$ be an open set, and let $f_1,f_2\in C^1(U)$. Prove that $$L_v(f_1f_2)=f_1L_vf_2+f_2L_vf_1$$

Suppose $f_1,f_2:U\rightarrow\mathbb{R}$. Let $p\in U$. We have $$L_v(f_1f_2)(p)=D(f_1f_2)(p)\cdot v=(f_1Df_2+f_2Df_1)(p)\cdot v=(f_1Df_2)(p)\cdot v+(f_2Df_1)(p)\cdot v.$$
The right-most term is the definition of $(f_1L_vf_2)(p)+(f_2L_vf_1)(p)$.
I just want to make sure: The notation $(f_1L_vf_2)(p)$ means $f_1(p)\cdot L_vf_2(p)$ (i.e. the product), right?

Yes,    $ f_1L_vf_2$ means pointwise  multiplication of functions. Hence, evaluating this product at $p$ amounts to multiplying $f_1(p)$ and $L_vf_2(p)$.
