On the Whitney-Graustein theorem and the $h$-principle. 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?
 A: 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!
