A ($\mathbb{Z}$-)grading on a ring $S$ is equivalent to an action of the group $\mathbb{G}_m$ on $U=\operatorname{Spec} S$. This is an exercise in algebra; the statement is that the symmetric monoidal category of comodules over the Hopf algebra $A[t,t^{-1}]$ is equivalent to the category of graded $A$-modules. The construction $\operatorname{Proj} S$ then corresponds to taking the quotient of $U\setminus U_0$ by the $\mathbb{G}_m$ action, where $U_0$ is the subscheme corresponding to the "irrelevant ideal". This is really just the familiar fact that $\mathbb{P}^n$ is the quotient of $\mathbb{A}^{n+1}\setminus 0$ by rescaling.

What happens when we take a product of two such schemes $U$ and $V$ with corresponding graded rings $S$ and $T$? I'm going to ignore $U_0$ and $V_0$ in this discussion to simplify notation (i.e., I'm going to pretend that $\operatorname{Proj} S$ is just $U/\mathbb{G_m}$). Well, $U\times V$ will naturally have an action of $\mathbb{G}_m\times \mathbb{G}_m$; this corresponds to the natural bigrading on $S\otimes T$. By taking the diagonal $\mathbb{G}_m\to\mathbb{G}_m\times\mathbb{G}_m$, we get an action of $\mathbb{G}_m$ on the product, and this corresponds to the "total" grading on $S\otimes T$. But if we mod out $U\times V$ by the diagonal action of $\mathbb{G}_m$, we don't get $U/\mathbb{G}_m\times V/\mathbb{G}_m$; to get that, we would need to mod out the entire action of $\mathbb{G}_m\times \mathbb{G}_m$. Alternatively, instead of modding out the whole action of $\mathbb{G}_m\times\mathbb{G}_m$ at once, we could first mod out the action of the antidiagonal $\alpha:\mathbb{G}_m\to\mathbb{G}_m\times\mathbb{G}_m$ given by $\alpha(t)=(t,t^{-1})$ and then mod out the action of the diagonal; this works since the antidiagonal and the diagonal together generate the whole group. What do we get when we mod out the antidiagonal? Well, it turns out that in this case the quotient will still be affine, and its coordinate ring can be found by taking invariants of the action on $S\otimes T$. If $x\in S_n$ and $y\in T_m$, then the the antidiagonal action of $t\in\mathbb{G}_m$ sends $x\otimes y$ to $t^{n-m}x\otimes y$, so the invariants are exactly $\bigoplus S_n\otimes T_n$. Thus the product $U/\mathbb{G}_m\times V/\mathbb{G}_m$ can be obtained as $\operatorname{Proj}(\bigoplus S_n\otimes T_n)$.