Proof. $\quad$ Consider the mapping
\begin{gather*}
f\colon V\to\mathbb{F}^m,
\end{gather*}
such that
\begin{gather*}
f(v)=\big(\phi_1(v),\dots,\phi_m(v)\big)
\end{gather*}
for all $v\in V.$ It is clear that $ \text{null} f=\text{null}(\phi_1)\cap\cdots \text{null}(\phi_m).$ Indeed, let $v\in \text{null}f.$ Then $f(v)=0,$ which implies that
\begin{gather*}
\big(\phi_1(v),\dots,\phi_m(v)\big)=0,
\end{gather*}
and thus $\phi_j(v)=0$ for $j=1,\dots,m.$ So we have $v\in \cap_{j=1}^m \text{null}(\phi_j).$ Hence we have proved $ \text{null} f\subset \cap_{j=1}^n \text{null} \phi_j.$ In the other direction of inclusion, suppose $v\in \cap_{j=1}^m \text{null}(\phi_j).$ Then $\phi_j(v)=0$ for all $j=1,\dots, m.$ Thus $f(v)=0,$ which implies that $v\in \text{null} f.$ Hence $\cap_{j=1}^m \text{null}(\phi_j)\subset \text{null} f.$ Therefore $\cap_{j=1}^m \text{null}(\phi_j)= \text{null} f.$
We then show that $\dim \text{range} f=m.$ Let $v_1,\dots, v_n$ be a basis of $V$ and $e_1,\dots, e_m$ be the standard basis of $\mathbb{F}^m.$
Then
\begin{gather*}
f(v_k)= \big(\phi_1(v_k),\dots, \phi_m(v_k)\big)
= \sum_{j=1}^{m}\phi_j(v_k)e_j.
\end{gather*}
Thus the matrix of $f$ with respect to the given bases is $A:=\big(\phi_j(v_k)\big)_{m\times n}.$ Therefore, by 3.117 and 3.118,
\begin{gather*}
\dim \text{range} f=\text{column}\text{ rank }{\mathcal{M}(f)}=\text{row rank } \mathcal{M}(f).
\end{gather*}
We claim that the list of rows of $A$ is linearly independent. Indeed, let $a_1,\dots, a_m\in\mathbb{F}.$ Suppose
$a_1A_{1,\cdot}+\cdots+a_mA_{m,\cdot}=0.$ That is,
\begin{gather*}
a_1\big(\phi_1(v_1),\dots, \phi_1(v_n)\big)+\cdots+a_m\big(\phi_m(v_1),\dots,\phi_m(v_n)\big)=0.
\end{gather*}
We have
\begin{gather*}
\big(a_1\phi_1(v_1)+\cdots+a_m\phi_m(v_1),\dots, a_1\phi_1(v_n)+\cdots+a_m\phi_m(v_n)\big)=0.
\end{gather*}
Thus
\begin{gather*}
a_1\phi_1(v_j)+\cdots+a_m\phi_m(v_j)=0,\qquad \text{for all $j=1,\dots, n.$}
\end{gather*}
Particularly,
\begin{gather*}
a_1\phi_1(v_1)+\cdots+a_m\phi_m(v_1)=0.\tag{1}
\end{gather*}
Because $\phi_1,\dots, \phi_m$ is linearly independent, from (1) it follows that $a_1=\cdots=a_m=0.$ Thus
the list of rows of $A$ is linearly independent. As a result, we have the row rank of $A$ is $m.$ Thus $\dim \text{range} f=m.$ Finally, by the Fundamental Theorem of Linear Maps, we have
\begin{align*}
&\dim\big( \text{null}(\phi_1)\cap\cdots\cap \text{null}(\phi_m)\big)=\dim \text{null} f\\
=&\dim V-\dim \text{range} f\\
=&\dim V-m.
\end{align*}