For $f \colon \mathbb{R}^n \to \mathbb{R}$ the gradient of $f$ is a vector field over $\mathbb{R}^n$ defined as $$\nabla f := \sum_{i=1}^n \frac{\partial f}{\partial x_i}\frac{\partial}{\partial x_i}.$$
How I can prove that $df_p(v) = \langle v, \nabla_p f \rangle$?
If $h \colon \mathbb{R}^n \to \mathbb{R}$ is another differential map, how to compute $[\nabla f, \nabla g]h?$
Since $$v = \sum_{k=1}^n v^k\frac{\partial}{\partial x^k}$$ then by definition of differential of a function you get \begin{align} df_p(v) & = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p}dx^i(v) = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p} \sum_{k=1}^n v^k dx^i\left(\frac{\partial}{\partial x^k} \right) \\ & = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p} \sum_{k=1}^n v^k \delta^i_k = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p} v^i = \langle v, \nabla_p f \rangle, \end{align} where $\langle \cdot, \cdot \rangle$ is the Euclidean metric on $\mathbb{R}^n$ and $\delta^i_k$ the Kronecker delta.
A hint for the second part: if $X = \sum_i X^i \partial_i$ and $Y = \sum_k Y^k \partial_k$ are two vector fields, then if $h$ is a smooth function \begin{align} [X,Y]h& = X(Yh)-Y(Xh) \end{align} with $Xh = \sum_i X^i \partial_i h$, and similarly for $Yh$.