Why was the definition defines as
$$ \forall \epsilon >0 \exists \delta \text{ such that if } 0<|x-a|< \delta \text{ then } |f(x)-L|< \epsilon$$
Rather than
$$\forall \epsilon >0 \exists \delta \text{ such that if } |f(x)-L|< \epsilon \text{ then } 0<|x-a|< \delta$$
The only reason I can think of is because $f(x)$ depends on $x$, but that isn't a satisfying answer since graphically it makes sense. That is as $f(x)$ gets closer to $L$, $x$ gets close to $a$. Secondly since it is for all $\epsilon$, it captures the idea of as "close as we want to $L$".
Side note What's confusing me here is that these two statements seem to capture the same idea which isn't the case.I'm hoping someone points out what's wrong with the second statement.
Consider the function $f$, defined by $$f(x) = \begin{cases} x+1, &x>0 \\ 0, &x=0 \\ x-1, &x<0\end{cases}$$
Clearly, it is discontinuous at $x=0$.
Now apply your second definition of continuity.
$\forall \varepsilon >0, \ \ \exists \delta>0$ such that $|f(x)-f(a)|<\varepsilon \implies |x-a|< \delta$.
I claim that for every $\varepsilon$, we can choose $\delta= \varepsilon$.
If $\varepsilon<1$, then $|f(x)-f(a)|<\varepsilon$ holds true only at $x=a$. So, $|x-a|<\delta$ is always true.
If $\varepsilon \geq 1$, then $|f(x)-f(a)|<\varepsilon$ is true when $x\in (-\varepsilon+1, \varepsilon-1) \subseteq (-\varepsilon,\varepsilon)$. So, $|x-a|<\delta$ is also true for this case.
Thus, your definition is satisfied for a discontinuous function.