Assume that $f_1, \ldots, f_k$ are linearly independent and extend them to a a basis $\{f_1, \ldots, f_n\}$ for the dual space $V^*$.
Let $\{b_1, \ldots, b_n\}$ be a basis for $V$ such that $\{f_1, \ldots, f_n\}$ is its dual basis.
Note that $\bigcap_{i=1}^k \ker f_i = \operatorname{span}\{b_{k+1}, \ldots, b_n\}$.
Indeed, obviously $b_j \in \ker f_i$ for $i = 1,\ldots, k$ and $j = k+1, \ldots, n$.
Conversely, let $x = \sum_{i=1}^n \alpha_ib_i \in \bigcap_{i=1}^k \ker f_i $. We have
$$0 = f_i(x) = \alpha_i, \quad\forall i = 1, \ldots, k$$
so $x = \sum_{i=k+1}^n \alpha_ib_i \in \operatorname{span}\{b_{k+1}, \ldots, b_n\}$.
Since $b_{k+1}, \ldots, b_n$ are linearly independent, we conclude $\dim \bigcap_{i=1}^k \ker f_i = n-k$.
If $f_1, \ldots, f_k$ are not linearly independent, we can assume that the first $f_1, \ldots, f_l$ are linearly independent, and that $f_{l+1}, \ldots, f_k$ are their linear combinations.
Then notice that
$$\bigcap_{i=1}^l \ker f_i \subseteq \ker f_j , \quad\forall j = k+1,\ldots, k$$
so $\bigcap_{i=1}^l \ker f_i = \bigcap_{i=1}^k \ker f_i$.
From the above proof follows that
$$\dim \bigcap_{i=1}^k \ker f_i =\dim \bigcap_{i=1}^l \ker f_i = n-l \ge n-k$$
Best Answer
Consider the linear map$$\begin{array}{rccc}\Psi\colon&V&\longrightarrow&\Bbb R^3\\&v&\mapsto&\bigl(f_1(v),f_2(v),f(v)\bigr).\end{array}$$Then $(0,0,1)\notin\Psi(V)$ and therefore there is a linear map $\varphi\colon\Bbb R^3\longrightarrow\Bbb R$ such that, for each $w\in\Psi(V)$, $\varphi(w)=0$ and that $\varphi(0,0,1)\ne0$. Then, for each $(a,b,c)\in\Bbb R^3$,$$\varphi(a,b,c)=\lambda_1 a+\lambda_2b +\lambda c,$$for some $\lambda_1,\lambda_2,\lambda\in\Bbb R$, with $\lambda\ne 0$ (otherwise, $\varphi(0,0,1)=0$). But then, for each $v\in V$,$$\lambda_1 f_1(x)+\lambda_2 f_2(v)+\lambda f(v)=0,$$and therefore$$f(v)=-\frac{\lambda_1}\lambda f_1(v)-\frac{\lambda_2}\lambda f_2(v).$$