Note: this post is about gaining an intuitive idea of why ultrafilter addition is defined the way it is. It assumes the reader already knows what an ultrafilter is, perhaps has encountered the extension of the semigroup structure of $\mathbb{N}$ to $\beta\mathbb{N}$ (the Stone-Czech compactification of the natural numbers), and is trying to understand that definition.
Addition seems like a straightforward concept. When I write $2+3$ or $m+n$ (when $m,n\in\mathbb{N}$), it is clear what I mean, and in the first case, the identification of $2+3$ with $5$ is also obvious. Also, when $n\in\mathbb{N}$ and $A\subset\mathbb{N}$, one can guess (correctly) that $A+n=\{a+n:a\in A\}$.
Perhaps slightly more complex is the case of '$A+B$', when $A$ and $B$ are sets of natural numbers. In this case, we add each element of $A$ to each element of $B$, in other words,
$$
A+B=\{a+b:a\in A, b\in B\}.
$$
Now, let us take things one step further. Rather than numbers, or sets of numbers, let us consider sets of sets of numbers - in particular, ultrafilters on $\mathbb{N}$. One might think ultrafilter addition can be easily defined in an analogous way to the way we added sets - by adding their elements. In the case of sets of naturals, the elements were natural numbers. In the case of ultrafilters, the elements are subsets of the naturals. In other words, one might be inclined to guess that ultrafilter addition, $+'$, is defined as $p+'q= \{ A+B:A\in p, B\in q \}$ .
However, this naive attempt does not give us the required result: the set $\{ A+B:A\in p, B\in q \}$ is not an ultrafilter. One can easily see that. Since we do not take zero to be a natural number, $\mathbb{N}$ contains no idempotent elements and no additive identity. Thus, there are no elements $n\in A\in p$ and $m\in B\in q$ such that $n+m=1$. Therefore, the set $\{ A+B:A\in p, B\in q \}$ does not contain $\mathbb{N}$ , as $1\in \mathbb{N} $, and we have just seen that there is no way to obtain a '$1$' through addition.
Let us change tactics. Since $1$ cannot be 'achieved' through addition, we can require the following for our new definition of addition: sets in $p+''q$ are those sets $C$ for which there is a set $B\in q$ such that $C-B\in p$. However, we again run into problems. What exactly does '$C-B$' mean? The obvious interpretation of $\{c-b:c\in C, b\in B\}$ might possibly include other integers than just the naturals. Thus, we might consider its restriction to the natural numbers: $C-B=\{c-b:c\in C, b\in B\}\cap\mathbb{N}$. Still, we have problems: namely, that these 'subtraction sets' are in a sense 'too big'. For example, if p is principal (say $\{2n\}\in p$ for some $n$) and the set of odd numbers $\{odd\}\in q$, then $\{2n\}+\{odd\}=\{2m+1:m\geq n\}$, in other words, the sum comprises of all the odd numbers greater than $2n$. However, by our definition of set subtraction, we would have $\{ 2m+1: m \geq n \}$ - $\{ odd \} = \{ even \}$ - definitely larger than the set we started with!
As a last attempt, we note that if instead we took $\{2m+1:m\geq n\}-\{2n\}$, we would have recovered the original set of odd numbers. So, instead of requiring that $C-B\in p$ for $B\in q$, we might need to consider a 'smaller' approach by requiring that $C-n\in p$ for $n\in B \in q$. In this way, we have reached the standard definition of ultrafilter addition,
$$
p\oplus q=\{C\subseteq\mathbb{N}:\{n\in\mathbb{N}:C-n\in p\}\in q\}.
$$
Addition seems like a straightforward concept. When I write $2+3$ or $m+n$ (when $m,n\in\mathbb{N}$), it is clear what I mean, and in the first case, the identification of $2+3$ with $5$ is also obvious. Also, when $n\in\mathbb{N}$ and $A\subset\mathbb{N}$, one can guess (correctly) that $A+n=\{a+n:a\in A\}$.
Perhaps slightly more complex is the case of '$A+B$', when $A$ and $B$ are sets of natural numbers. In this case, we add each element of $A$ to each element of $B$, in other words,
$$
A+B=\{a+b:a\in A, b\in B\}.
$$
Now, let us take things one step further. Rather than numbers, or sets of numbers, let us consider sets of sets of numbers - in particular, ultrafilters on $\mathbb{N}$. One might think ultrafilter addition can be easily defined in an analogous way to the way we added sets - by adding their elements. In the case of sets of naturals, the elements were natural numbers. In the case of ultrafilters, the elements are subsets of the naturals. In other words, one might be inclined to guess that ultrafilter addition, $+'$, is defined as $p+'q= \{ A+B:A\in p, B\in q \}$ .
However, this naive attempt does not give us the required result: the set $\{ A+B:A\in p, B\in q \}$ is not an ultrafilter. One can easily see that. Since we do not take zero to be a natural number, $\mathbb{N}$ contains no idempotent elements and no additive identity. Thus, there are no elements $n\in A\in p$ and $m\in B\in q$ such that $n+m=1$. Therefore, the set $\{ A+B:A\in p, B\in q \}$ does not contain $\mathbb{N}$ , as $1\in \mathbb{N} $, and we have just seen that there is no way to obtain a '$1$' through addition.
Let us change tactics. Since $1$ cannot be 'achieved' through addition, we can require the following for our new definition of addition: sets in $p+''q$ are those sets $C$ for which there is a set $B\in q$ such that $C-B\in p$. However, we again run into problems. What exactly does '$C-B$' mean? The obvious interpretation of $\{c-b:c\in C, b\in B\}$ might possibly include other integers than just the naturals. Thus, we might consider its restriction to the natural numbers: $C-B=\{c-b:c\in C, b\in B\}\cap\mathbb{N}$. Still, we have problems: namely, that these 'subtraction sets' are in a sense 'too big'. For example, if p is principal (say $\{2n\}\in p$ for some $n$) and the set of odd numbers $\{odd\}\in q$, then $\{2n\}+\{odd\}=\{2m+1:m\geq n\}$, in other words, the sum comprises of all the odd numbers greater than $2n$. However, by our definition of set subtraction, we would have $\{ 2m+1: m \geq n \}$ - $\{ odd \} = \{ even \}$ - definitely larger than the set we started with!
As a last attempt, we note that if instead we took $\{2m+1:m\geq n\}-\{2n\}$, we would have recovered the original set of odd numbers. So, instead of requiring that $C-B\in p$ for $B\in q$, we might need to consider a 'smaller' approach by requiring that $C-n\in p$ for $n\in B \in q$. In this way, we have reached the standard definition of ultrafilter addition,
$$
p\oplus q=\{C\subseteq\mathbb{N}:\{n\in\mathbb{N}:C-n\in p\}\in q\}.
$$