Prove $\bigcap_{i=1}^k\ker(f_i)\subset \ker(f)\iff f\in {\rm span}(f_1,...,f_k) $ Let $V$ a $\mathbb{K}$-vector space of finite dimension $n$, with $\{f_1,...,f_n\}$ a set linearly independent of $V^*$ and $f\in V^*$.
Prove $\bigcap_{i=1}^k\ker(f_i)\subset \ker(f)\iff f\in {\rm span}(f_1,...,f_k).$
($\Leftarrow$) Let $f\in {\rm span}(f_1,...,fk)$ then exists $\alpha_1,...,\alpha_k$ such that $f=\alpha_1f_1+...+\alpha_kf_k$.
Let $f\in\bigcap_{i=1}^k\ker(f_i) $ then $f\in \ker(f_1),...,f\in \ker(f_k)$.
By hypothesis we have if $f\in {\rm span}(f_1,...,fk)$ then exists $\alpha_1,...,\alpha_k$ such that $f=\alpha_1f_1+...+\alpha_kf_k$
Here I'm a little stuck. Can someone help me?
($\Rightarrow$) I do'nt know how to prove this part. Help me, if you can. I will be very grateful.

You start well, but soon make a mistake: the statement $f\in\bigcap_{i=1}^k\ker(f_i)$ is wrong, because the kernels are subspaces of $V$ and $f\in V^*$.
What you have to prove is

If $f\in\operatorname{Span}(f_1,\dots,f_k)$ then $\ker(f)\supset\bigcap_{i=1}^k\ker(f_i)$.

Suppose $f\in\operatorname{Span}(f_1,\dots,f_k)$; then $f=\alpha_1f_1+\dots+\alpha_kf_k$ for some scalars $\alpha_1,\dots,\alpha_k$. If $x\in\bigcap_{i=1}^k\ker(f_i)$, then $f_i(x)=0$, for $i=1,\dots,k$ and therefore
$$
f(x)=\alpha_1f_1(x)+\dots+\alpha_kf_k(x)=0
$$
proving that $x\in\ker(f)$.
Now let's try the converse:

If $\ker(f)\supset\bigcap_{i=1}^k\ker(f_i)$ then $f\in\operatorname{Span}(f_1,\dots,f_k)$.

Suppose $\ker(f)\supset\bigcap_{i=1}^k\ker(f_i)$. Write
$$
f=\alpha_1f_1+\dots+\alpha_kf_k+\alpha_{k+1}f_{k+1}+\dots+\alpha_nf_n
$$
which is possible because $\{f_1,\dots,f_n\}$ is a basis of $V^*$.

Lemma. Every basis of $V^*$ is the dual of a basis $\{e_1,\dots,e_n\}$ of $V$.

Once accepted this lemma (for the proof, see https://math.stackexchange.com/a/1772676/62967), we have $\{e_1,\dots,e_n\}$ such that
$$
f_i(e_j)=\begin{cases} 1 & i=j \\ 0 & i\ne j \end{cases}
$$
In particular, $e_{k+1},\dots,e_n\in\bigcap_{i=1}^k\ker(f_i)$, so
$$
f(e_j)=0,\quad j=k+1,\dots,n
$$
and therefore, for $j=k+1,\dots,n$,
\begin{align}
0=f(e_j)&=\alpha_1f_1(e_j)+\dots+\alpha_kf_k(e_j)+
\alpha_{k+1}f_{k+1}(e_j)+\dots+\alpha_jf_j(e_j)+\dots+\alpha_nf_n(e_j) \\
&=0+\dots+0+0+\dots+\alpha_j\cdot1+\dots+0 \\
&=\alpha_j
\end{align}
Hence $\alpha_j=0$ for $j=k+1,\dots,n$ and finally
$$
f=\alpha_1f_1+\dots+\alpha_kf_k\in\operatorname{Span}(f_1,\dots,f_k)
$$
