Help in understanding what is going on in $A[G]$ Ok, so Im given a group $G$ and a ring $A$, and define:
$$A[G]=\left\{\sum_{g \in G} f(g)  g : f:G \to A, \text{ such that $f$ has finite support}  \right\}$$
Define the sum $(+)$:
$$\sum_{g \in G} f(g)  g + \sum_{g \in G} h(g)  g = \sum_{g \in G} (f(g)+h(g))  g$$
And multiplication $(\cdot)$:
$$\sum_{\alpha \in G} f(\alpha)  \alpha \cdot \sum_{\beta \in G} h(\beta)  \beta = \sum_{g\in G}\left(\sum_{\alpha \beta = g} f(\alpha)h(\beta)\right)  g$$
And then $(A, +, \cdot)$ is a ring.
Now, Im trying to calculate $Z(A[G])$, and Ive got a suggestion that says to first show that if $\sum_{g \in G} f(g)  g \in Z(A[G])$ then $f(g) \in Z(A)$ for every $g\in G$. 
So I take  $\sum_{g \in G} f(g)  g \in Z(A[G])$ and then for any  $\sum_{g \in G} h(g)  g \in A[G]$, we have that
$$\sum_{\alpha \in G} f(\alpha)  \alpha \cdot \sum_{\beta \in G} h(\beta)  \beta =\sum_{\beta \in G} h(\beta)  \beta \cdot \sum_{\alpha \in G} f(\alpha)  \alpha$$
So I thought, ok, given $x\in A$ lets take $h:G \to A$ given by $h(1_{G})=x$ and $h(g)=0$ for every $g \in G-\{1\}$, in order to show that $f(g) \in Z(A)$ for every $g \in G$. So here comes my misunderstanding: For this given $h$, 
$$\sum_{\alpha \in G} f(\alpha)  \alpha \cdot \sum_{\beta \in G} h(\beta)  \beta = \sum_{\alpha \in G} f(\alpha)  \alpha \cdot (x \cdot 1_{G} + \sum_{g \in G-\{1\}} 0 \cdot g)$$
But $0 \in A$ and $g \in G$ and (I've have not studied Modules until now) and therefore, In principle I don't know how to multiply $0 \cdot g$. In fact I don't know how to multiply any $a \in A$ times any $g\in G$. So what's the issue? I don't need that? Where am I lost?
 A: Think of the $g$ in $\sum_{g \in G} f(g) g$ as placeholders, the same way as the $x^j$ act as placeholders in a polynomial's expression $a_nx^n + \dotsb + a_1x^1 + a_0x^0$. This should make it easier to see that $$\sum_{g \in G} f(g) g = \sum_{g \in G} h(g) g$$ if and only if $$f(g) = h(g)\quad \text{for every } g \in G.$$
Using your choice of $h$, for any given $x \in A$ you get
$$
\sum_{\alpha \in G} f(\alpha)\alpha \cdot \sum_{\beta \in G} h(\beta)\beta
= \sum_{\alpha \in G} \big(f(\alpha)x\big) \alpha
= \sum_{\alpha \in G} \big(xf(\alpha)\big) \alpha
= \sum_{\beta \in G} h(\beta)\beta \cdot \sum_{\alpha \in G} f(\alpha)\alpha
$$
which by the remark above means that $f(g)x = xf(g)$ for every $g \in G$.

Following the OP's comment below, let's look a bit more in detail at how one can compute those products. By definition
$$
\sum_{\alpha \in G} f(\alpha)\alpha \cdot \sum_{\beta \in G} h(\beta)\beta
= \sum_{g \in G} \left(\sum_{\alpha\beta=g} f(\alpha) h(\beta)\right) g
$$
so let's compute the sum in brackets for a given $g \in G$. Now, consider some $\alpha,\beta \in G$ such that $\alpha\beta = g$, and suppose that $\beta \neq 1_G$. Then by your choice of $h$ we have $h(\beta) = 0$, thus $f(\alpha) h(\beta) = 0$. Since the sum in brackets is computed in $A$, it follows that
$$
\sum_{\alpha\beta=g} f(\alpha) h(\beta) = f(g)h(1_G) + \sum_{\substack{\alpha\beta=g\\\beta\neq1_G}} 0 = f(g)x.
$$
Similarly for the other product.

On the other hand, note that when you have some concrete elements of $A[G]$ you can compute products in a simpler way. Indeed, for ease of notation let's agree that, when writing out an element of $A[G]$, we will omit all terms with coefficient $0$. Then the finiteness condition means that every non-zero element of $A[G]$ is of the form $a_1 g_1 + \dotsb + a_n g_n$ for some $g_1,\dotsc,g_n \in G$ and $a_1,\dotsc,a_n \in A$ non-zero. Furthermore,
$$
\begin{align}
(a_1g_1+\dotsb+a_ng_n)(b_1h_1+\dotsb+b_mh_m) &= (a_1b_1)(g_1h_1) + \dotsb + (a_1b_m)(g_1h_m)\\&+(a_2b_1)(g_2h_1) + \dotsb + (a_nb_m)(g_nh_m).
\end{align}
$$
If this looks a lot like a product of polynomials, you are not mistaken. The same definitions for sum and product in $A[G]$ still work when $G$ is a monoid instead of a group, and if you take $G = \{X^n : n \in \Bbb{Z}_{\geq 0}\}$ you recover the ring of polynomials $A[X]$.
With this in mind, let's revisit the previous product:
$$
\sum_{\alpha \in G} f(\alpha)\alpha \cdot \sum_{\beta \in G} h(\beta)\beta
= \left(\sum_{\alpha \in G} f(\alpha)\alpha\right) (x1_G)
= \sum_{\alpha \in G} (f(\alpha)x)(\alpha1_G)
= \sum_{\alpha \in G} (f(\alpha)x)\alpha.
$$
