Directional derivative and composite functions

We have a function $$f(p_0)$$ with $$p_0 \in \mathbb{R}^n$$, and the vector $$\vec{v}$$ with $$||\vec{v}|| = 1$$. The derivative of $$f(p)$$ to $$\lambda$$ in the point $$p = p_0 + \lambda \vec{v}$$, calculated for $$\lambda = 0$$, is called directional derivative along $$\vec{v}$$ and is indicated with the symbol $$\left( \dfrac{\partial f}{\partial v} \right)_{p_{0}}$$

From the composite function derivation rules, must be

$$\left[ \dfrac{d}{d \lambda} f(p_0 + \lambda \vec{v})\right]_{\lambda = 0} = f'_{x_1}(p_0) \cdot v_1 +... f'_{x_n}(p_0) \cdot v_n$$

Although I have consulted several books, this passage is still not clear to me, especially how the function $$f(p_0 + \lambda \vec{v})$$ should be a composite function. Can you help me?

Well, $$f(p)$$ is a function of $$p$$ to begin with, but here $$p=p(\lambda)=p_0 + \lambda \vec v$$ is a function of $$\lambda$$. So you get a composite function which in the end depends on $$\lambda$$. And its derivative is found using the multivariable chain rule.
You can write $$f(p)$$ as a function of all the components of $$p_0$$:$$f(p)=f(p_0+\lambda\vec v)=f(p_{0,1}+\lambda v_1,p_{0,2}+\lambda v_2,...,p_{0,n}+\lambda v_n)$$ We now have $$n$$ components, so $$f'(p_0+\lambda\vec v)=\frac{d(p_{0,1}+\lambda v_1)}{d\lambda}\partial_{p_{0,1}+\lambda v_1}f+...+\frac{d(p_{0,n}+\lambda v_n)}{d\lambda}\partial_{p_{0,n}+\lambda v_n}f\\=f'_{p_{0,1}+\lambda v_1}(p_0+\lambda\vec v)v_1+...+f'_{p_{0,n}+\lambda v_n}(p_0+\lambda\vec v)v_n$$