As suggested in a comment, it's hard to give a reasonable answer in this generality; if you show us a specific "contrived" proof people may be able to supply some motivation for the contrivance. But in this generality the question comes very close to "How do I learn to prove stuff?", and while good answers to that question exist they tend to be book-length.
Contrivances in General
Of course a big part of the answer to "How can I learn to prove stuff?" is "Learn the standard techniques in the area". Note
If a contrived trick gets used several times it has by definition become a standard technique.
Contractions
So. How did someone come up with that proof of IFT? Probably she or he knew this:
If you have to show that $f(x)=a$ has a solution it's often useful to find a different function $g$ such that you need to show $g$ has a fixed point.
This is so because there are various simple ways to show $g$ has a fixed point, for example by showing $g$ is a contraction. The first person to give that proof of IFT had seen several examples of solving equations by finding fixed points, so he or she decided to try it. It worked, thus adding to the list of examples where that particular trick is useful.
IFT
Some years ago I put IFT on my personal list of things I know how to prove. Not because I'd learned a proof. I never really feel I understand anything unless I've figured it out for myself; I convinced myself that I could prove IFT by changing it to a problem about fixed points.
Let's see if I can still do that. The actual theorem has a lot of technicalities, but the essence of it, the hard part, is this:
Theorem (most of IFT). Suppose $f:\Bbb R^n\to\Bbb R^n$ is $C^1$, $f(0)=0$ and $Df(0)$ is invertible. There exists $\delta>0$ such that if $|y|<\delta$ then there exists $x$ with $f(x)=y$.
The proof is immediate from the heuristic above plus one other "trick", which we can "motivate" (since I came up with the trick myself I just have to explain how I thought of it).
Motivation. Fix $y$ with $|y|$ small. We need to show that $$f(x)=y$$has a solution. This is the same as showing that $g$ has a fixed point, where $$g(x)=f(x)-y+x.$$
So it's enough to show that $g$ is a strict contraction. By a little calculus, it would be enough to show there is a convex neighborhood $C$ of the origin such that the operator norm of the derivative satisfies $$||Dg(x)||\le\lambda<1\quad(x\in C).$$But $$Dg(x)=Df(x)-I$$(that is, $$Dg(x)h=Df(x)g-h.)$$Hmm. Since $Dg$ is continuous, this would show that $||Dg(x)||$ was small for small $x$ if we had $Df(0)=I$. Hmm... Aha. The condition $Df(0)=I$ is actually WLOG!
Proof. Or actually not so much a proof as an explanation of the main step in the proof; I leave it to you to insert the quantifiers and the logic to convert this into something I'd actually accept as a proof from a student.
First, replacing $f$ by $f\circ\psi$ for an appropriate diffeomorphism $\psi$, we can assume that $$Df(0)=I.$$ (For example, if $T=Df(0)$ let $F(x)=f(T^{-1}x)$; showing $f(x)=y$ has a solution is the same as showing $F(x)=y$ has a solution.)
Fix $y$ and define $g$ as above. Since $Df$ is continuous there exists $\epsilon>0$ such that $$||Df(x)-I||\le 1/2\quad(|x|<\epsilon).$$So $$||Dg(x)||<1/2\quad(|x|\le\epsilon)..$$ So $g$ is a strict contraction, hence $g$ has a fixed point.
I bet that's more or less the same as the proof in the book, since it's the obvious (sorry) way to try to prove it using a contraction. (If it looks very different I conjecture that's because he didn't start with the "WLOG $Df(0)=I$"; if so try rewriting the proof in the book starting with $Df(0)=I$ tthat will simplify a lot of the formulas, probably making it all look more like what's above.)
Note. For a minute it seemed like that was a proof that $f$ is surjective, which of course is nonsense, because I didn't see where we used the fact that $y$ is small.
But duh. What's above reads as though the theorem was "Any strict contraction has a fixed point", which is also nonsense. The actual theorem is of course
Theorem(Banach) If $X$ is a complete metric space and $g:X\to X$ is a strict contraction then $g$ has a fixed point.
So it's enough to prove this:
There exists $r>0$ such that if $0<\rho<r$ there exists $\delta>0$ such that if $|y|<\delta$ then $g(X)\subset X$, for $X-\overline{B(0,\rho)}$.
Hint: $$|g(x)|\le|y|+|g(0)-g(x)|\le |y|+c|x|.$$
