Why does creating a model show consistency? As per the title, why does the ability to generate a model from axioms prove they are consistent?
I would just like to mention that one generally does not construct a model of a theory $T$ from the axioms of $T$ itself.  Instead, one uses external means to produce a structure that one can show is a model of $T$.

Here is where Ben's answer comes into play, although in my opinion it obfuscates matters: if you construct a model of $T$ using some other theory $T^\prime$, then what you have really done is shown that the consistency of $T^\prime$ implies the consistency of $T$.  There are times where this distinction is important.  For example, from the axioms ZFC of set theory one can construct a model of Peano Arithmetic, PA.  It is unknown whether ZFC is actually consistent, though widely believed, and therefore we cannot immediately conclude that PA is consistent, though this is also widely believed.

But let's assume that we have constructed a model $M$ of $T$ using methods that are above reproach.  For example, $M$ might be a finite $\mathcal{L}$-structure of specific size $n$.  The claim that $T$ is consistent then comes down to two facts about the underlying logic.

Fact 1: If $M$ is a structure in some language $\mathcal{L}$, then $M \models \varphi$ and $M \models \neg \varphi$ cannot both be true for every $\mathcal{L}$-sentence $\varphi$.  (In fact, exactly one of $M \models \varphi$ or $M \models \neg \varphi$ holds.)

Fact 2: If $T$ is a theory in a language $\mathcal{L}$ and $\varphi$ is a $\mathcal{L}$-sentence such that $T \vdash \varphi$, then $M \models \varphi$ for all models $M$ of $T$.

Now, if $\varphi$ were an $\mathcal{L}$-sentence such that $T \vdash \varphi$ and $T \vdash \neg \varphi$, then by Fact 2 we conclude that $M \models \varphi$ and $M \models \neg \varphi$, but this contradicts Fact 1!
Creating a model doesn't show that a theory is consistent. What it shows is that, if we can create a model of theory A within theory B, then we know that if A is inconsistent, B must be inconsistent as well.

For example, say A is elliptic geometry. At one time, people thought that A was inconsistent, and they tried really really hard to prove that it was inconsistent. But let B be Euclidean geometry. We can construct a model of A in B, by taking A's points to be points on a sphere, identifying points that lie at each other's antipodes, and taking lines to be great circles. This shows that if A was inconsistent, then B would as well.

In most cases, there is no hope of proving B's consistency, due to Godel's theorem. However, it often happens that B is a theory that people don't think is likely to be inconsistent, e.g., ZFC set theory or Peano arithmetic. Therefore A has a model in B, everybody realizes that it's a waste of time to worry about inconsistencies in A.