It comes down to whether or not you want $K(X)$ to be a group. If all you care about is isomorphism classes of vector bundles, that's a perfectly good monoid. People wouldn't call the monoid $K$-theory because $K$-theory of stuff generally means you've performed some kind of group-completion of a monoid or a category or something.
So there's two ways to respond to your question. The simplest would be "why don't vector bundles have inverses?" which on the most superficial level has the answer not all vector bundles are $0$-dimensional.
Another way to respond would be, what if we performed the formal group-completion of the monoid of isomorphism classes of vector bundles? Well, you get the same thing. General nonsense says you have a map from the group completion to $K(X)$ defined your way, since bundles over reasonable spaces have complementary bundles.
Is this roughly an answer to your question?
edit in response to your comment: Say $X$ is a finite-dimensional CW-complex, and $\epsilon$ is a finite-dimensional vector bundle over $X$. Then there is a classifying map for $\epsilon$, this is a map $X \to G_{\infty,k}$ where $k$ is the dimension of $\epsilon$. $G_{\infty,k}$ is the Grassmannian of $k$-dimensional subspaces of $\mathbb R^\infty$. A basic theorem about $G_{\infty,k}$ is that any map of a finite-dimensional CW-complex into $G_{\infty,k}$ is homotopic to a map into $G_{m,k}$ for $m$ perhaps very large. $G_{m,k}$ is canonically isomorphic to $G_{m,m-k}$ (take orthogonal complements). The bundle classified by the corresponding map $X \to G_{m,m-k}$ is the complementary bundle to $\epsilon$, denote it $\epsilon'$. By design, $\epsilon \oplus \epsilon'$ is a trivial bundle as it's the pull-back of the tangent bundle to $\mathbb R^m$. The fact that maps $X \to G_{\infty,k}$ are homotopic to maps $X \to G_{m,k}$ has many proofs -- one is from the Schubert cell decomposition of $G_{m,k}$ together with cellular approximation.
That's the idea. I believe complementary bundles exist for more than bundles over finite-dimensional spaces but that's the most convenient argument that comes to mind (have to run off to teach a class now).
2nd edit: Let $\mathcal V(X)$ be the group completion of the monoid $\mathcal M(X)$ of isomorphism classes of finite-dimensional vector bundles over $X$. Let $K(X)$ be the stable isomorphism classes of vector bundles over $X$. $K(X)$ is a group for a fairly broad class of spaces, by the arguments above. So by group-completion formal nonsense there is a map $\mathcal V(X) \to K(X)$. The question is whether this map has an inverse or not. Given an element $[\alpha] \in K(X)$, $\alpha \in \mathcal M(X)$ is a vector bundle over $X$ and is well-defined up to stable equivalence. To $\alpha$ there is an associated element of $\mathcal V(X)$ and the question is, is the map
$$K(X) \to \mathcal V(X)$$
well-defined, where you map $[\alpha]_{K(X)} \longmapsto [\alpha]_{\mathcal V(X)}$?
Ah, okay, I see the problem now. This isn't well-defined. But things are off only by a tiny thing. $\mathcal V(X)$ contains a copy of the integers as a direct-summand. This is because $\mathcal M(X) \to \mathbb Z$ given by taking the dimension of a bundle is a homomorphism of monoids (using addition in $\mathbb Z$). There is a map back $\{0,1,2,\cdots \} \to \mathcal M(X)$ given by taking trivial bundles over $X$. So in $\mathcal V(X)$ this gives a splitting $\mathcal V(X) = \mathbb Z \oplus \overline{\mathcal V(X)}$ where $\overline{\mathcal V(X)}$ is the "reduced" group-completion of isomorphism types of vector bundles over $X$. There is a well-defined map $K(X) \to \overline{\mathcal V(X)}$ since you can think of $\overline{\mathcal V(X)}$ as being $\mathcal V(X)$ modulo trivial bundles.
Does that make sense now? I haven't thought about this stuff in much detail for years.
First of all: you shouldn't give up on problems after 30 minutes. Take a break, try a different problem, maybe wait a few days and try again -- you'll gain a lot more from the problem if you struggle and solve it yourself. Having access to solutions can be helpful, but you don't want to find yourself relying on them. (There's a phrase that gets thrown around a lot: "If you can't solve a problem then there's an easier problem you can't solve; find it").
Baby/Blue Rudin is a great book for an introduction to the basics of analysis (beyond one-variable "advanced calculus"). After that I'd suggest looking at the 'Lectures in Analysis' series written by Elias Stein and Rami Shakarchi (Stein was actually Terrence Tao's advisor). These books cover introductory Fourier analysis, complex analysis, measure theory, and functional analysis. Along the way the authors expose you to all kinds of in-depth and enlightening applications (including PDEs, analytic number theory, additive combinatorics, and probability). Of all the analysis textbooks I've looked at, I feel like I've gained the most from the time I've spent with Stein and Shakarchi's series -- these books will expose you to the "bigger picture" that many classical texts ignore (though the "classics" are still worth looking at).
I've skimmed through parts of Terrence Tao's notes on analysis, and these seem like a good option as well (though I looked at his graduate-level notes, I don't know if this is what you're referring to). I've always gotten a lot out of the expository stuff written by Tao, so you probably can't go wrong with the notes regardless. If you feel like you need more exercises, don't be afraid to use multiple books! Carrying around a pile of books can get annoying, but it's always helpful to see how different authors approach the same subject.
Best Answer
"Complex Topological K-Theory" by Efton Park is a pretty decent introduction to topological K-Theory, but I'd actually go with Karoubi's book.
There is also a chapter on K-Theory in John Peter May's "A Concise Course in Algebraic Topology", which is in my opinion the best text on algebraic topology currently available, including some references and recommended further reading.