Intuition for maxterms
Indeed, maxterm is the concept dual to minterm. As you mention,
$$
f(x) = \bigvee_{y\colon f(y)=1} \text{"$x=y$"} = \bigvee_{y\colon f(y)=1} \bigwedge_{i=1}^n \text{"$x_i=y_i$"}.
$$
Here $\vee$ is OR (your addition), $\wedge$ is ADD (your multiplication), the input is $x = x_1,\ldots,x_n$, and "$x_i=y_i$" is either $x_i$ (if $y_i$ is True) or $\bar{x}_i$ (if $y_i$ is False).

Similarly, we have
$$
f(x) = \bigwedge_{y\colon f(y)=0} \text{"$x \neq y$"} = \bigwedge_{y\colon f(y)=0} \bigvee_{i=1}^n \text{"$x_i \neq y_i$"}.
$$
This time "$x_i\neq y_i$" is $x_i$ if $y_i$ is False, and $\bar{x}_i$ if $y_i$ is True.



When the function is monotone, we can improve on the first representation:
$$
f(x) = \bigvee_{\text{$y$ minterm}} \text{"$x \geq y$"} = \bigvee_{\text{$y$ minterm}} \bigwedge_{i\colon y_i=1} x_i.
$$
Here "$x \geq y$" means $x_i \geq y_i$ for all $i$, and a minterm is a satisfying assignment of $f$ which is minimal in the sense that changing any 1 to 0 changes it to an unsatisfying assignment.

We can define maxterms analogously – maximal unsatisfying assignments, and then
$$
f(x) = \bigwedge_{\text{$y$ maxterm}} \text{"not $x \leq y$"} = \bigwedge_{\text{$y$ maxterm}} \bigvee_{i\colon y_i=0} x_i.
$$