If you are in a hurry and this question still has caught your interest, please jump directly to the last proposition, where my question lies.
Throughout this question I am going to identity $\mathbb{S}^1$ and $[0,1]/\partial[0,1]$. Therefore, when I will talk about mappings from $\mathbb{S}^1$ to $\mathbb{R}^2$, I will consider maps $f\colon[0,1]\rightarrow\mathbb{R}^2$ such that $f(0)=f(1)$.
Let $I(\mathbb{S}^1,\mathbb{R}^2)$ be the set of immersions of $\mathbb{S}^1$ into $\mathbb{R}^2$, that is the set of $C^1$-mappings from $\mathbb{S}^1$ to $\mathbb{R}^2$ such that their derivatives do not vanish. My goal is to prove the well-known:
Theorem. (Whitney-Graustein) The turning number gives a bijection from $\pi_0(I(\mathbb{S}^1,\mathbb{R}^2))$ to $\mathbb{Z}$.
For sake of clarity, by now let $X:=C^0(\mathbb{S}^1,\mathbb{R}^2\setminus\{(0,0)\})$. Inspired by the Gromov's $h$-principle, I introduced the following map: $$J\colon\left\{\begin{array}{ccc}I(\mathbb{S}^1,\mathbb{R}^2)&\rightarrow&X\\f&\mapsto&f'\end{array}\right..$$
I claim that one has the following:
Theorem. The map $J$ induces a well-defined bijection $$\pi_0(J)\colon\left\{\begin{array}{ccc}\pi_0(I(\mathbb{S}^1,\mathbb{R}^2))\rightarrow\pi_0(X)\\ [f]_0\mapsto [f']_0\end{array}\right..$$
I have already prove the well-definedness and the following:
Proposition. Let $f\in X$, there exists $g\in I(\mathbb{S}^1,\mathbb{R}^2)$ and $H\colon\mathbb{S}^1\times[0,1]\overset{C^0}{\rightarrow}\mathbb{R}^2\setminus\{(0,0)\}$ such that $H(\cdot,0)=f$ and $H(\cdot,1)=g'$.
Proof. On request. $\Box$
Which has direct corollary $\pi_0(J)$ being surjective. Hence, I am left to establish the following:
Proposition. Let $g_1,g_2\in I(\mathbb{S}^1,\mathbb{R}^2)$ such that there exists $H\colon\mathbb{S}^1\times [0,1]\overset{C^0}\rightarrow\mathbb{R}^2\setminus\{(0,0)\}$ such that $H(\cdot,0)={g_1}'$ and $H(\cdot,1)={g_2}'$. Then, there exists $F\colon\mathbb{S}^1\times [0,1]\overset{C^1}{\rightarrow}\mathbb{R}^2$ such that $F(\cdot,0)=g_1$, $F(\cdot,1)=g_2$ and for all $t\in [0,1],F(\cdot,t)\in I(\mathbb{S}^1,\mathbb{R}^2)$.
Proof. My idea is to integrate the homotopy $H$, that is introducing: $$F(x,t):=\int_0^xH(u,t)\,\mathrm{d}u-x\int_0^1H(u,t)\,\mathrm{d}u.$$ The removed corrective term is here to ensure that for all $t\in [0,1]$, $F(0,t)=F(1,t)$, that is for the well-definedness of $F(\cdot,t)$ on $\mathbb{S}^1$. Notice that one has $F(\cdot,0)=g_1-g_1(0)$ and $F(\cdot,1)=g_2-g_2(0)$. Therefore, if for all $t\in [0,1]$, $F(\cdot,t)\in I(\mathbb{S}^1,\mathbb{R}^2)$, I am almost done. However, it is not clear and wrong in all generality, that the following quantity is nonzero: $$\frac{\mathrm{d}}{\mathrm{d}x}F(x,t)=H(x,t)-\int_{0}^1H(u,t)\mathrm{d}u.$$ That is where I am stuck. $\Box$
Question. If $H(x,\cdot)$ is non-constant and belongs to $\mathbb{S}^1$, I am done. Indeed, $\displaystyle\int_0^1H(u,t)\mathrm{d}u$ will lie in the interior of the unit disk. However, it is not clear to me that I can boil down my problem to this case and doing a naive radial on $H(x,\cdot)$ homotopy does not seem to help in anything.
If the latter proposition is true, I am done with the injectivity of $\pi_0(J)$ and with the Whitney-Graustein theorem. Indeed, I will have the following commutative diagramm, where all arrows are bijections:
$$\require{AMScd}\begin{CD} \pi_0(I(\mathbb{S}^1,\mathbb{R})) @>\textrm{turning number}>> \mathbb{Z}\\ @VV\pi_0(J)V @AA\deg A\\ \pi_0(X) @>\textrm{str. def. retract}>> \pi_1(\mathbb{S}^1)\end{CD}$$
Any enlightenment will be greatly appreciated. If my approach proving the last proposition is plain wrong, could you provide me some other thoughts to manage my way toward the proof?
You're pretty close to having all of the details. First of all, you can assume (by homotopies of the initial $g_1$ and $g_2$) that $g_1$ and $g_2$ both have total length $1$ and are parametrized by arc length. In this case, the ranges of $g_1^\prime$, $g_2^\prime$, and $H$ are all $S^1$, which resolves one of the issues in your final question.
Now, for any fixed $t \in (0,1)$, $H(x,t) : S^1 \to S^1$ is homotopic to $g_1^\prime$ and $g_2^\prime$, and therefore has the same degree as those two maps. In particular, if the degree of those two is nonzero, then $H(x,t)$ cannot be constant for any fixed $t$, and your argument works. It remains to show that if the degrees are zero, you can choose $H$ so that $H(x,t)$ is nonconstant for every fixed $t$.
If the maps $H(\cdot,0),H(\cdot,1):S^1 \to S^1$ have degree $0$, we can lift to maps $h_0,h_1 : S^1 \to \mathbb{R}$ (with respect to the usual covering $\mathbb{R} \to S^1 : t \mapsto e^{it}$). Note that this would not be possible for other degrees, since the image would not be a loop. We can also homotope the initial $g_0$ and $g_1$ so that $h_0(x) = h_1(x)$ for all $x\in[0,\varepsilon]$ for some small $\varepsilon$, and also so that $h_0$ and $h_1$ are nonconstant on this interval. Now define $h_t$ as the straight-line homotopy from $h_0$ to $h_1$:
$$h_t(x) = (1-t)h_0(x) + th_1(x),$$
and finally,
$$H(x,t) = ( \cos(h_t(x)), \sin(h_t(x)) ),$$
which is a homotopy with all the properties you need. $\square$
If you haven't done so, check out the book "h-Principles and Flexibility in Geometry" by Geiges. It's an excellent, readable intro to h-principle which includes this proof!