Try to prove the following:
Two paths $\gamma_1,\gamma_2\colon I\to X$ from $p$ to $q$ are homotopic relative the endpoints if and only if the loop $\gamma_1*\overline{\gamma_2}$ at $p$ is null-homotopic (relative the basepoint).
Here $\overline{\gamma_2}$ denotes the reversed path of $\gamma_2$ and $*$ denotes concatenation of paths.
From this it then follows that the homotopy class of a path $\gamma\colon I\to S^1$ relative the endpoints consists of all paths $\gamma'$ such that the loop $\gamma*\overline{\gamma'}$ has winding number $0$.
The idea of going from a homotopy $H\colon I\times I\to X$ with $H(0,-)=\gamma_1*\overline{\gamma_2}$, $H(1,-)= p$ and $H(-,0)=H(-,1)=p$ to a homotopy from $\gamma_1$ to $\gamma_2$ is shown in the following picture. You can transform $\gamma_1$ into $\gamma_2$ like the blue lines show, keeping $\gamma(0)=p$ and $\gamma(1)=q$ at all times.
Let me sum up what I wrote in comments.
So how to understand what is the connection between the homotopy equivalence of topological spaces and paths they correspond to?
I think this question reveals the main flaw in your reasoning. Spaces do not correspond to any path. You confuse a path with its image. Path is just a continuous function $f:[0,1]\to X$. And space is a space. And indeed, two paths can be homotopic, while their images are not.
The simplest example is as follows: let $\lambda:[0,1]\to X$ be any path. Then $\lambda\cdot\overline{\lambda}$ is always homotopic to the constant loop, regardless of what its image is (which is equal to the image of $\lambda$). Assume furthermore that there is a surjective path $[0,1]\to X$, i.e. $X$ is a Peano space. Does this imply that every Peano space is contractible? Of course not. $S^1$ (or any sphere) is a non-contractible Peano space.
The opposite situation is also possible, i.e. two paths have homotopic images, while they are not homotopic. For example consider $S^1$ as a subset of complex numbers of norm $1$. Then consider two loops $S^1\to S^1$, $z\mapsto z$ and $z\mapsto z^2$ (I use $S^1$ in domain instead of $[0,1]$ since loops over $[0,1]$ correspond to continuous functions over $S^1$). It is well known these are not homotopic (Hopf theorem) but their images are not only homotopic, they are literally equal (both to full $S^1$).
All in all: do not think about paths as their images. This is sometimes useful, but generally is wrong.
For me a path is a way of moving a point over a space in time. The image is not that important, because we can visit a single place multiple times, we can go backwards, etc. So "how" we move matters more, image itself is not enough. Now if $X$ is additionally locally compact Hausdorff space then a homotopy between two paths $[0,1]\times[0,1]\to X$ is the same as path $[0,1]\to C([0,1], X)$, where $C([0,1],X)$ denotes the space of all continuous functions $[0,1]\to X$ (with compact-open topology). And that's how I personally think about homotopies: it's a path but in the space of continuous functions. Sort of higher level path. Of course you have to be careful, this intuition doesn't work for non-locally compact spaces. Still, it covers lots of everyday examples (e.g. manifolds and locally finite CW complexes).
On the pictures below $f$ and $f_1\cdot f_2$ are path homotopic, and yet the corresponding spaces are not homotopy equivalent
What exactly are those corresponding spaces? Here are my guesses (from the most probable to the least probable):
- The space $X$ is $\mathbb{R}^3\backslash S^1$ where $S^1$ is embedded into $\mathbb{R}^3$, say by adding $0$ at the last coordinate. Then we consider two greyed paths $f$ and $f_1\cdot f_2$. Indeed, their images are even homeomorphic, but these are not homotopic inside our $X$. Note the importance of $X$. If you add the missing black circle they are again homotopic, without modyfing those paths. And so being homotopic is not an internal property of paths, it heavily depends on the ambient space.
- The space in question is the complement of union of black and grey circle. In that case indeed, two spaces can be homotopic while their complements are not. The simplest example is $X=[0,1]$, $A=\{0\}$, $B=\{\frac{1}{2}\}$. In this case $A$ and $B$ are clearly even homeomorphic, but $X\backslash A$ is not homotopy equivalent to $X\backslash B$ because the first one is connected while the second one is not.
- The space in question are grey paths, and we actually look at them as knots, and we consider ambient isotopy. Indeed, two spaces can be homotopic, but not ambient isotopic. In fact every two knots are even homeomorphic, because by definition, a knot is an embedding of $S^1$ into $\mathbb{R}^3$. I'm putting this interpretation here, since you've mentioned homeomorphisms in comments. Note that in general paths are rarely homeomorphisms, e.g. every path $S^1\to S^1$ is homotopic to $z\mapsto z^n$ for precisely one $n$ (again: Hopf theorem), but these are homeomorphisms only for $n=\pm 1$. So in a sense most paths $S^1\to S^1$ are not homeomorphisms. In fact your own $f_1\cdot g\cdot\overline{g}\cdot f_2$ path is not a homeomorphism.
Either way, you have to be precise. It is unclear what happens in your second drawing (the first one is sort of ok). What spaces I am working with? What are those paths? What are their formulas? What $x_0$ is? And since it is different in both cases why $x_0$ matters? Finally what homotopy theory I'm working with? Once you make everything as precise as possible, I'm sure you will be able to answer your own questions.
Best Answer
A homotopy between paths is just a map out of $[0,1]^2$ satisfying some properties. The labels on these diagrams say, what values the map takes in the target space $X$.
Let’s discuss the second square in your second image. By convention we treat the $x$-component as path-parameter and the $y$-component as homotopy-parameter. We want that the homotopy preserves endpoints. This means that no matter where in the homotopy-parameter we are we want the path to start at $x_0$ and end in $x_1$. This is the reason why the left and right edge of the square are labeled as such. In explicit formulas we want: $H(0,y) = x_0$ and $H(1,y)=x_1$.
The second condition in this example is that it should be a homotopy from $c_{x_1}\ast\gamma$ to $\gamma$. This is why the bottom and top edges are labeled as such. In formulas $H(x,0) = c_{x_1}\ast\gamma(x)$ and $H(x,1)=\gamma(x)$.
Now the obvious question is what do we do at an arbitrary point $(x,y)$? One sensible thing would be that when fixing the homotopy-parameter $y$ we get the path which traverses $\gamma$ in the interval $[0,\frac{1}{2} + y\frac{1}{2}]$ and then traverses the constant path $c_{x_1}$ the remaining $[\frac{1}2+y\frac{1}2,1]$. This is what the diagonal line is representing with the label $x_1$ reminding us, that (at least for fixed $y$) the paths are compatible at this line.
This discussion leads us to the formula $$H(x,y) = \left \{\begin{array}{ll} \gamma((2-y)x) & x \leq \frac{1}2 + y \frac{1}2\\ c_{x_1}((2-y)x) & \text{else} \end{array}\right.$$ which I hope I did not mess up.
To get explicit formulas out of the homotopy diagrams is a bit cumbersome, mostly because it involves making case distinctions and annoying reparametrizations. I do not know of a general recipe besides the fact that it usually is a matter of linear interpolations. The point of these diagrams is however that they make rather clear that one can find a homotopy. The idea of above homotopy is: take the identity homotopy of $\gamma$ and the identity homotopy of $c_{x_1}$, glue them together along $x_1$ and reparametrize such that the domain is the unit square.