I've gone through the similar post Modulo of a negative number . But that post is not about proof and I'm asking for the proof in general.
This question is another follow up question of my previous question. The question is how can we prove that $(a-b) \mod N = a \mod N + ((-b) \mod N)$.
The comments are getting complicated.
The main point of working modulo $N$ is to make multiples of $N$ irrelevant, so we define $a\equiv b \bmod N$ if and only if $N\mid a-b$ or equivalently if there is an integer $r$ with $a=b+rN$.
The relation $\equiv$ defined in this way is easily seen to be an equivalence relation which divides the integers $\mathbb Z$ into $N$ equivalence classes.
It is easily seen that $(a+kN)+(b+rN)=a+b+(k+r)N$
and $(a+kN)(b+rN)=ab+(ar+bk+krN)N$
So addition and multiplication respect equivalence.
One way of doing arithmetic modulo $N$ is to choose a representative element of each equivalence class. Often this is the remainder on division by $N$, giving representatives $0, 1, 2 \dots N-1$ (I'll call this the standard set). Sometimes it is more convenient to choose the residues of least absolute value (with $N/2$ taken for the relevant class if $N$ is even).
As regards negative numbers we have $-b\equiv N-b\equiv 2N-b \dots$ and this will eventually give a positive residue.
Though it makes sense to choose a set of representatives of the classes - for example, the standard set - and do our arithmetic using these, it is easy to see that these cannot be closed under addition or multiplication* (take the residue of greatest absolute value and add it to itself, or square it). The result may have to be adjusted by a multiple of $N$ to give the required residue. The same is true of subtraction.
You can either try to work over the integers and have to adjust your results by multiples of $N$ to get equality, or you can work with equivalences throughout.
If in doubt go back to the definition and work over $\mathbb Z$ using arbitrary multiples of $N$ - the language of equivalence does the same job whilst simplifying the expressions and the arithmetic.
*For multiplication there is an exception: the residues $\{0,1\}$ are closed under multiplication modulo $2$