How does one prove that if $f$ and $g$ are linear functionals on $V$ such that $h=fg$ is also a linear functional, then either $f=0$ or $g=0?$

For any $v,w\in V$, \begin{eqnarray} f(v)g(v)+f(w)g(w)&=&h(v)+h(w)=h(v+w)=f(v+w)g(v+w)\\\\ &=&f(v)g(v)+f(w)g(w)+f(v)g(w)+f(w)g(v). \end{eqnarray} We conclude that $$ f(v)g(w)+f(w)g(v)=0,\ \ v,w\in V. $$ In particular, $f(v)g(v)=0$ for all $v\in V$. Now let $e_1,\ldots,e_n$ be a basis of $V$. If $f\ne0$, then there exists $i$ with $f(e_i)\ne0$.

We have $0=f(e_i)g(e_i)$, so $g(e_i)=0$. For any other $j$, $$ 0=f(e_i)g(e_j)+f(e_j)g(e_i)=f(e_i)g(e_j); $$ as $f(e_i)\ne0$, we get that $g(e_j)=0$ for all $j$, i.e. $g=0$.

Another way to complete the story (using your starting point that $h=0$, and therefore $V=\mathrm{Ker}\,f\cup \mathrm{Ker}\,g$), is to prove that the union of two proper subspaces of $V$ cannot ever be all of $V$. This is true over any field. Here is a (rather big) hint for how to prove this:

Let $W,W'$ be proper subspaces of $V$. If one is contained in the other, their union is not all of $V$ because they are both assumed proper. If neither is contained in the other, then there exists $x$ in $W$ but not in $W'$ and $x'$ in $W'$ but not in $W$. What can you say about $x+x'$?