I'm trying to filling the details about the construction of the long exact sequence in K-theory. I'm using the notation of Hatcher's book (pages 52-53).
here is a related question, even thought it's a step ahead. I want to construct such sequences.
According to Hatcher and other authors, the ingredients are two. I'll denote by X a compact Hausdorff (non empty) space, and $A$ a closed subspace of $X$. $S$ is the suspension, and $C$ is the Cone of a space.
fact 1 given $$ A \xrightarrow{i} X \xrightarrow{q} X/A$$ s.e.s then $$ \tilde{K}(X/A) \xrightarrow{q^*} \tilde{K}(X) \xrightarrow{i^*} \tilde{K}(A)$$ is s.e.s
the other ingredient is
fact 2 if $A$ is contractible, there is a bijection $$\text{Vect}(X/A) \approx \text{Vect}(X)$$
So given the sequence 
I've proved that $$ \tilde{K}(X/A) \xrightarrow{q^*} \tilde{K}(X) \xrightarrow{i^*} \tilde{K}(A)$$ is s.e.s (fact 1) $$ \tilde{K}(SA) \xrightarrow{q_2^*} \tilde{K}(X \cup CA) \xrightarrow{i^*} \tilde{K}(X)$$ is s.e.s (fact 1 again) and so on...
by $q_2 \colon (X \cup CA) \to (X \cup CA)/X \approx SA$ I mean the quotient map the collapse $X$ to a point. The homeomorphism is evident.
So how can I glue this two s.e.s? I don't know how to use fact 2 to glue. In my opinion this is the way to do, I hope I don't have misunderstood something.
I took sometimes to sketch the diagrams needed. I will not write down a complete proof, I just sketch the main passages and then is up to you to fill the details (easy, don't worry).
So first of all, note that fact 2 let us define a isomorphism between $\tilde{K}(X/A)$ and $\tilde{K}(X)$. In fact it preserve the direct sums (note that the pullback of the quotient is its inverse and it preserve sums)
Then we change notation:
With this notation, we have that $$ X_{n+2}/CX_n \approx SX_{n-1}$$
We can write down this sequence of short exact sequences:
Where the $i$'s denote the obvious inclusion, the $q$'s the quotients. The dotted lines should help you visualize the situation and our next step. Thanks to our observations and notation, we can add the following arrow $$ b_n \colon X_n \to X_{n-1}/X_{n-2}$$ which represent the map that collapse the last attached cone to the space $X_n$, obtaining in fact $X_{n-1}/X_{n-2}$. So we can sketch this new diagram:
We make a claim CLAIM 1 The diagram above is indeed commutative.
Sketch of proof: the following equality holds $$ b_n \circ i_{n+1} \equiv q_n$$ (it should be kind of obvious)
Using fact 1 we have a sequence of short exact sequences obtained "turning the arrow". We need to GLUE them together, as you said, using fact 2. Let us define $$ f_n := (b_n^*)^{-1} \circ q_{n+1}^*$$ We are able to draw this commutative diagram:
The definition of $f$ holds thanks to fact 2 applied to $b_n$ (cones are contractible) and our observation. So we need to prove the exactness of the sequence with uniform arrow (not the dotted ones).
How to prove exactness: just make an intensive use of this three facts:
1) $b_n^*$ is bijective
2) the short sequences of fact 1 are exact
3) $i_{n+1}^* \circ b_n^* \equiv (b_n \circ i_{n+1})^* \equiv q_n^*$
$$\times \times \times \times $$
I apologize in advance for the -not so organized- answer. I copy-pasted some of my diagrams (source available if needed), because according to me, was the best option.