The modulo operator $a\mapsto a\%c$ sends $\Bbb Z\to\{0,1,\cdots,c-1\}.$ (This is the domain and range.)
Now, is $\%c$ an additive function? That is, does $(a+b)\%c=a\%c+b\%c$? Notice we would have to have the latter two add to a number $. Also, if $a,b$ are already in $\{0,\cdots,c-1\}$ they are un-affected and we would have $(a+b)\%=a+b$. But the sum of two numbers $ is not necessarily itself less than $c$; what if $a=c-1$ and $b=1$ for instance? Thus it is not additive.
Notice that $(x\pm c)\%=x\%c$ for any $x$. That is, it is a periodic function. Moreover, if we subtract the remainder $x\%c$ from $x$ we will end up with a multiple of $c$. Symmetrically, if we subtract this multiple of $c$ from $x$ we obtain the remainder $x\%c$! So, if this "multiple of $c$" is $nc$ for $x=a$ and then $mc$ for $x=b$, by the periodicity condition we have
$(a+b)\%c=(a-nc+b-mc)\%c=(a\%c+b\%c)\%c=(a+b\%c)\%c.$
Remember that $a$ is already in the range of $\%c$ so $a\%c=a$. This is the best we can say at this level of generality on the variable $b$. We cannot remove the final $\%c$; remember our $b=1,a=c-1$ example. (This is a rather longwinded reasoning process that more or less mirrors the perhaps more efficient deduction possible using modular arithmetic and basic elementary number theory.)
Yes, it is true that $(a_1+\cdots+a_n)\%c=(a_1\%c+\cdots+a_n\%c)\%c \tag{$*$}$
This can be proven just as with the $n=2$ case. Use modular reduction to say that
$a_i=m_ic+a_i\%c$
for each $i=1,\cdots,n$. Then by periodicity we have
$(a_1+\cdots+a_n)\%=\big((a_1-m_ic)+\cdots+(a_n-m_nc)\big)\%c=(a_1\%c+\cdots+a_n\%c)\%c \tag{$\circ$}.$
Using this, we have
$(\underbrace{a+\cdots+a}_b)\%c=\big(\underbrace{a\%c+\cdots+a\%c}_b)\%c=\big(b\cdot(a\%c)\big)\%c. \tag{$\bullet$}$
By setting each $a_i=a$ and $n=b$ in the formula $(\circ)$.
Furthermore, setting $b=\ell c+b\%c$, by periodicity again we have
$\big(b\cdot(a\%c)\big)\%c=\big(b(a\%c)-\ell(a\%c)c\big)\%c=\big((b-\ell c)(a\%c)\big)\%c=\big((b\%c)(a\%c)\big)\%c.$