A formal proof that a sum of infinite series is a series of a sum? I feel confused when dealing with ininities of any kind. E.g. the next equation is confusing me.
$$\displaystyle\sum^\infty_{n} (f_1(n) + f_2(n)) = \displaystyle\sum^\infty_{n_1=1} f_1(n_1) + \displaystyle\sum^\infty_{n_2=1} f_2(n_2)$$
How do people deal with such a special object as infinity with such an ease (it makes sense in general though)? Is there any axiomatic theory for that? I ask because I saw people relying on this when calculating the exact value of a series.
And last but not least: does my question sound tautological? Better delete it then.

You have not quite stated a result fully. We state a result, and then write down the main elements of a proof. You should at least scan the proof, and then go to the final paragraph. 
We will show that if $\sum_{i=1}^\infty f(i)$ and $\sum_{j=1}^\infty g(j)$ both exist, then so does $\sum_{k=1}^\infty (f(k)+g(k))$, and
$$\sum_{k=1}^\infty (f(k)+g(k))=\sum_{i=1}^\infty f(i)+\sum_{j=1}^\infty g(j).$$
Let $\sum_{i=1}^\infty f(i)=a$ and $\sum_{j=1}^\infty g(j)=b$. 
We want to show that $\sum_{k=1}^\infty (f(k)+g(k))=a+b$. 
Let $\epsilon \gt 0$. By the definition of convergence, there is an integer $K$ such that if $n \gt K$ then 
$$\left|\sum_{i=1}^\infty f(i)-a\right|\lt \epsilon/2.$$
Similarly, there is an integer $L$ such that if $n \gt L$ then 
$$\left|\sum_{j=1}^n g(j)-b\right|\lt \epsilon/2.$$
Let $M=\max(K,L)$. Then, from the two inequalities above, and the Triangle Inequality, it follows that if $n \gt M$, we have
$$\left|\sum_{k=1}^n (f(k)+g(k))-(a+b)\right|\lt \epsilon.$$
This completes the proof.
If you do not have experience with arguments like the one above, it may be difficult to understand. But there is one important thing you should notice. After the initial statement of the result, there is no more mention of "infinity." All subsequent work is with finite sums. The existence and value of an expression like $\sum_{n=1}^\infty h(n)$ is defined purely in terms of finite sums.  
Remark: It is perfectly possible for the sum on your left-hand side to exist, while the sums on the right do not. A crude example would be $f(i)=1$ for all $i$, and $g(j)=-1$ for all $j$. But one can prove that if any two of the sums exist, then the third one does.  
