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?
 A: 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!
A: Given a language $\mathsf{L}$, we want to construct of a model from a set of given axioms. Let us call the $\mathsf{L}$-sentences $A =\{\varphi_{1},...,\varphi_{n}\}$ the axioms. An $\mathsf{L}$-theory (call it $\mathbf{T}$) is a set of $\mathsf{L}$-sentences, so we have that $A$ is an $\mathsf{L}$-theory and so is any other set of $\mathsf{L}$-sentences. It is defined that $\mathcal{M}$ is a model for $\mathbf{T}$ if $\mathcal{M} \vDash \psi$, for all $\mathsf{L}$-sentences $\psi \in \mathbf{T}$.
When a theory $\mathbf{T}$ is inconsistent, it means that there is at least one $\psi \in \mathbf{T}$ such that $\mathbf{T} \vdash \psi \wedge \mathbf{T} \vdash \neg \psi$. Yet, our definition of a model $\mathcal{M}$ for an $\mathsf{L}$-theory $\mathbf{T}$ is exactly that $\mathcal{M} \vDash \psi$, for all $\psi \in \mathbf{T}$.
Hence it follows that for our set of axioms $A = \{\varphi_{1},...,\varphi_{n}\}$, that if we can construct a model $\mathcal{M}$ for $A$ it means that $\mathcal{M} \vDash \varphi_{i}$, where $1 \leq i \leq n$. Therefore, it must necessarily be the case that if we can construct a model $\mathcal{M}$ for our set of axioms $A$, that $A$ is consistent, otherwise we would not be able to construct such a model $\mathcal{M}$.
If you have any other interesting questions about model theory feel free to ask, I'm glad to help! Let me know if you need any clarification about the terminology or you aren't acquainted with what some of the symbols are.
