Given this definition:
and the following theorem:
This theorem does not mention anything about whether $f$ is injective or not, and the theorem states that if $a \in f(U)$ is a regular value of $f$ , then $f^{-1} (a) $ is a regular surface in space.
But $f^{-1} (a)$ is a set of points such that $f(x,y,z) = a$ , my question is, what if $f$ is injective? Then $f^{-1} (a)$ must be a unique single value, and that single point cannot possibly be a regular surface, so what's wrong with my understanding?
Nothing wrong there. It's just that differentiable functions from open subsets of $\Bbb R^3$ to $\Bbb R$ are never injective. In fact, you've already argued that for an injective differentiable function $f:U\to\Bbb R$ no value $a\in f(U)$ can be regular. Therefore, for all $x\in U$, $f^{-1}(f(x))$ must contain some $y$ such that $d_yf=0$. Since $f^{-1}(f(x))=\{x\}$, it must be $y=x$. Specifically, $d_xf=0$ for all $x\in U$. You should have somewhere (I recommend reading your book, but here is an instance) the result that $d_xf=0$ for all $x$ implies that $f$ is locally constant (and therefore not injective).