Following Hatcher, we define Chern classes $c_i$ taking vector bundles $E \to X$ to some equivalence class $c_i(E) \in H^{2i}(X; \mathbb Z)$ But when we define the Chern character, $$\operatorname{ch}(E) := \dim E + \displaystyle\sum_{k>0} \frac{s_k(c_1(E), c_2(E), \ldots, c_k(E))}{k!},$$ where $s_k$ are the Newton polynomials, we suddenly end up with an element in $\displaystyle\prod_{k\geq 0} H^k(X; \mathbb Q)$.
I do not understand this. Why different coefficient rings? What difference does it make?
The Chern character satisfies $\text{ch}(E \otimes F) = \text{ch}(E) \text{ch}(F)$ and $\text{ch}(E \oplus F) = \text{ch}(E) + \text{ch}(F)$, as well as the usual axioms of naturality and its value on, $\tau_1 \to \Bbb{CP}^1$.
The Chern character is the unique natural transformation to $\prod H^{2i}(X;R)$ satisfying these properties. There is no integer-cohomology-valued natural transformation that satisfies them. Explaining why is outside the scope of this answer, but the reason we introduce the rationals is that we must.
As for why we suddenly end up with something with $\Bbb Q$ coefficients, you just divided by $k!$, and most integers are not divisible by $k!$. Given an integer cohomology class $x$ and an integer $n$, you can always make sense of $\frac 1n x \in H^*(X;\Bbb Q)$, but this expression need not make sense in $H^*(X;\Bbb Z)$ (first: an element $y$ with $ny = x$ need not exist; second, if the cohomology ring has $n$-torsion, it is not unique).