This is rigorous; the loop described by image $(1)$ is homologous to the "loop" described by image $(4)$ in the singular homology of $X:=\Bbb R^2\setminus\{0\}$ and we could convert the diagrams into a formal proof of this (taking as a given that there exist continuous paths that "look like" those pictures).
To pass from $(1)$ to $(2)$, you observe that the loops there are path-homotopic thus homologous. Why? A based homotopy is of those two loop is a continuous $H:I\times I\to X$. By partitioning $I\times I$ into two copies of $\Delta^2$ along the diagonal, restricting $H$ gives two simplices (of the singular simplicial set of $X$) $a,b:\Delta^2\to X$ where $b$ has been chosen so that its diagonal is oppositely oriented to $a$. $a+b$ is an element of $C_2$ whose formal boundary is the sum $f-g+u-v=f-g$ where $f,g:\Delta^1\to X$ are parametrisations of our loop and $u,v$ are the constant maps to the loop basepoint, so in fact $u=v$ and these cancel in the space of formal linear combinations that is $C_1$. Therefore, $f-g$ is a boundary element of $C_1$ (it is in the image of the formal boundary operator $C_2\to C_1$); it follows that $f=g+(f-g)$ is equivalent to $g$ in the first singular homology group. That gives us power to say $(1)$ is homologous to $(2)$.
To figure out $(2)\to(4)$ we first need to note that if $f:x\to y,g:y\to z$ are paths in $X$ then the ($1$-simplex represented by the) path composite $gf$ is homologous to the formal sum $f+g$. This is true just because there's a hollow triangle with edges $f,g,gf$ and we can easily extend this to a $2$-simplex (without needing "more points", so there is no issue with crossing zero etc.); just take $\Delta^2\twoheadrightarrow\Delta^1\overset{gf}{\longrightarrow}X$ where the first arrow projects onto the relevant edge.
So we can decompose $(2)$ as homologous to the formal sum of every "piece", seeing it as a piecewise function. In $(3)$, a triangle is drawn and shaded at the bottom; two of its edges are "pieces" from $(2)$. The triangle is a pictorial representation of the fact there is a continuous $\Delta^2\to X$ whose formal boundary equals $a+b-c$ in:
$a+b-c$ is then a boundary element of $C_1$, so subtracting this from the $1$-simplex representing our loop from $(2)$ doesn't change its homology; and, $(2)$ is just (homologous to) a formal sum of (the other pieces) with $a$ and $b$. The homologous result of this subtraction is (by reattaching all the "pieces") then homologous to $(3.5)$:
Which is just $(2)$ with the edges $a$ and $b$ "cancelled out" and replaced with $c$. The main point is that it is rigorous to perform this cancellation when working modulo boundaries.
Now, this is homologous to $(4)$ + the $1$-simplex representing walking around the edges of the shaded square. By similar reasoning - following the shaded diagram - or by noting this last loop is nullhomotopic in $X$, this $1$-simplex is homologous with zero. Therefore, it is rigorous to say $(1)$ is homologous to $(2)$ is homologous to $(4)$, taking certain basics as a given.
