You seem to misunderstand the grading a bit. Look at $X=CP^2$ and the $K$-theory class $H-1$, where $H$ is the Hopf bundle. Then $ch(H-1)= 1+z+\frac{z^2}{2}-1$, with $z$ a generator of $H^2 (CP^2;Z)$. Let $i=2$. Each map $Y \to CP^2$ from an $(i-1)$-dimensional $Y$ is nullhomotopic, since $CP^2$ is simply connected. Therefore, $H-1$ belongs to $K^{0}_{(2)} (CP^2)$.
On the other hand, the component of $ch(H-1)$ in $H^4(CP^2)$ does not refine to an integral class.
What I can offer you is an explanation of the fact that the $i$th component of the Chern character of an element $x \in K^{0}_{(2i)} (X)$ is integral. Assume, for simplicity, that $X$ is a finite CW complex. Let $f: X \to BU$ be a classifying map for the $K$-theory class $x$. The assumption says that $f|_{X^{(2i-1)}}$ is nullhomotopic. By the homotopy extension property, $f$ factors through the $(2i-1)$-connected complex $Y=X/X^{(2i-2)}$.
By the universal coefficient theorem in cohomology, it is enough to prove that $ch_i$ evaluates to an integer on each integral homology class $a \in H_{2i}(Y, Z)$. By Hurewicz, it is enough to verify that for each $g: S^{2i} \to Y$, $\langle g^{\ast} ch_i, [S^{2i}]\rangle$ is integral.
The arguments so far reduce the question to the following result:
Theorem: let $V \to S^{2n}$ be a complex vector bundle. Then $\langle ch(V), [S^{2n}] \rangle$ is integral.
This is a highly nontrivial result and it is not really elucidated by taking appropriate definitions of the Chern classes. In fact, you need either Bott periodicity or the index theorem for Dirac operators. Here are the two alternative arguments.
Bott periodicity theorem (this is done in, say, somewhere in the book ''Topology of Lie groups'' by Mimura and Toda). You have to prove that the Chern character $K^{0}_{cpt} (C^n) \to H_{cpt}^{ev} (C^n;Q)$ takes integral values. Since the Chern character is multiplicative, this amounts to proving that the Bott class $b \in K^{0}_{cpt} (C)$ has $ch(b)$ integral (this reduction uses Bott periodicity!). But $ch(b)\in H^{2}_{cpt} (C)$ is the generator, by direct computation!
Index theorem: The Hirzebruch class of $S^{2n}$ is $\mathcal{L}=1$. Take the signature operator $D_V$, twisted by the bundle $V$. The index is
$$
ind (D_V)= \langle \mathcal{L}(TS^{2n}) ch (V); [S^{2n}]\rangle= \langle ch (V); [S^{2n}]\rangle
$$
and the left-hand side is integral. Note that the version of the index theorem that can be proved using the heat kernel is enough.
