You can phrase it another way. Suppose $S = I(V\times W) \setminus (I(V) + I(W))$ is nonempty. Among $f\in S$, pick $f$ so that it has an expression $f = \sum_{i=1}^n f_i g_i$ with $f_i \in k[x_1,\dots,x_n]$ and $g_i \in k[x_{n+1},\dots,x_{n+m}]$ for all $1\le i \le n$ so that $n$ is minimal, i.e. there is no other $h\in S$ with such an expression but with fewer terms.
Since $f \notin I(W)$, we must have that some $g_i \notin I(W)$. So there is some $b\in W$ with $g_i(b) \ne 0$, and we assume without loss of generality that $i=1$. Now, as in the post you refer to, there is $p\in I(V)$ so that
$$f_1 = \frac{p}{g_1(b)} - \frac{g_2(b)}{g_1(b)} f_2 - \dots - \frac{g_n(b)}{g_1(b)}f_n.$$
Now we have
$$f = \sum_{i=1}^n f_i g_i = \left(\frac{p}{g_1(b)} - \frac{g_2(b)}{g_1(b)} f_2 - \dots - \frac{g_n(b)}{g_1(b)}f_n\right)g_1+ \sum_{i=2}^n f_i g_i$$
$$= \frac{p\cdot g_1}{g_1(b)} + \sum_{i=2}^n f_i \left(g_i-\frac{g_i(b)}{g_1(b)}g_1\right).$$
Now, if $f-\frac{p\cdot g_1}{g_1(b)} \in I(V) + I(W)$, then $f \in I(V) + I(W)$ too, since $\frac{p\cdot g_1}{g_1(b)} \in I(V)$ (the expanded ideal). So $h = f-\frac{p\cdot g_1}{g_1(b)} \notin I(V) + I(W)$. But clearly $h \in I(V+W)$, and $h$ has the form in the first paragraph with fewer terms than $f$ had. This contradicts the minimality of the number of terms in the expression of $f$.
You can always convert these arguments which decrease some integer bounded-below-parameter of some set of objects you want to prove don't exist (i.e., said set is empty) with a minimality argument like this one. They are much cleaner, in my opinion. It's like you're arguing with someone who thinks they exist and instead of saying "give me any of them and I'll show you you're wrong but it might take a while", you cut to the end and just say "give me the one I want and I can show you you're wrong right away".
