Questions about amortised analysis
Let $a_i$ be the amortized costs of operation $i$, $c_i$ be the actual costs of operation $i$, and $D_i$ the data structure after operation $i$. The amortized costs of an operation  are defined as
$$a_i:=c_ i+ \Phi(D_i) -\Phi(D_{i-1}).$$
You usually assume the following


The potential is always positive, that is $\forall\colon i \Phi(D_i)\ge 0$,
In the beginning the potential is 0, that is $\Phi(D_0)=0$.


These two assumptions are not always necessary but they are commonly used and make the life easier. Now you can consider the potential at costs that have already been paid by previous operations.

To justify this definition consider the accumulated costs of all operations. We get the following telescoping sum
$$
\sum_{i=1}^n a_i = \sum_{i=1}^n (c_i +\Phi(D_i) -\Phi(D_{i-1})) = \Phi(D_n) -\Phi(D_{0}) + \sum_{i=1}^n c_i \ge \sum_{i=1}^n c_i. 
$$
So you see that the sum of the amortized costs exceeds the total actual costs. Hence the amortized costs give an upper bound.

I have not seen the use of the $c$ in your formula. But the $c$ can be simply added to the potential function $\Phi$. So think of the $c$ as a factor that weights the potential, and be redefining the function $\Phi$ you can get rid of it. 

So to answer your concrete questions. $a(i)$ denotes the amortized costs for the $i$th operation, that is the actual costs in a sequence of operation charged to operation $i$. The $c$ is some positive factor you can pick. And the $a(i)$s are not constant in general (they are in your example) - you don't have to show anything for the $a(i)$ costs, the values $a(i)$ are the result of your analysis.

Since you seem to speak German, you might also have a look at my German class notes to the potential function.