Finding a non-constant homotopy between two degree zero maps $\mathbb{S}^1\rightarrow\mathbb{S}^1$. For technical reasons and in the context of a wider proof, I would like to prove the following:

Lemma. Let $f\colon\mathbb{S}^1\rightarrow\mathbb{S}^1$ and $g\colon\mathbb{S}^1\rightarrow\mathbb{S}^1$ be two continuous map of degree zero, there exists a homotopy $H\colon\mathbb{S}^1\times[0,1]\overset{C^0}{\rightarrow}\mathbb{S}^1$ such that $H(\cdot,0)=f$, $H(\cdot,1)=g$ and which is non-constant over  $\mathbb{S}^1\times\{t\}$.

If $f$ and $g$ have the same base point, then it is no big deal to find a homotopy between $f$ and $g$. However, this is not necessarily the case and the point is that I need it to be non-constant when restricted to each $\mathbb{S}^1\times\{t\},t\in [0,1]$.
The idea is to consider $\theta_f\colon[0,1]\rightarrow\mathbb{R}$ and $\theta_g\colon [0,1]\rightarrow\mathbb{R}$ be continuous lifts of $f$ and $g$ through the map $x\mapsto(\cos(2\pi x),\sin(2\pi x))$ such that $\theta_f$ and $\theta_g$ are equal and non-constant on $[0,\varepsilon]$ for a fixed $0<\varepsilon\leqslant 1$.
Which homotopies should I perform on $f$ and $g$ to do so? I would like it to be explicit.
Any enlightenment will be greatly appreciated!
 A: As I am only interested in the case where $f$ and $g$ are both derivatives of immersions of $\mathbb{S}^1$ into $\mathbb{R}^2$, I will solely prove the following:

Proposition. Let $g_1,g_2\colon\mathbb{S}^1\rightarrow\mathbb{R}^2$ be two immersions of $\mathbb{S}^1\rightarrow\mathbb{R}^2$ such that ${g_1}',{g_2}'\colon\mathbb{S}^1\times\mathbb{S}^1$ and both have degree zero, there exists a continuous $H\colon\mathbb{S}^1\times [0,1]\rightarrow\mathbb{S}^1$ such that $H(\cdot,0)={g_1}'$, $H(\cdot,1)={g_2}'$ and for all $t\in[0,1]$, $H_{\vert\mathbb{S}^1\times\{t\}}$ is non-constant.

I will identify $\mathbb{S}^1$ with the quotient $\mathbb{R}/\mathbb{Z}$.
Let $p$ be the universal cover of $\mathbb{S}^1$ defined by: 
$$p\colon\left\{\begin{array}{ccc}
\mathbb{R} & \rightarrow & \mathbb{S}^1\\
x & \mapsto & (\cos(2\pi x),\sin(2\pi x))
\end{array}\right..$$
For $i\in\{1,2\}$, there exists $\theta_i\colon\mathbb{R}\rightarrow\mathbb{R}$ a continuous lift of ${g_i}'$ through $p$. Since, ${g_i}'$ have degree zero, $\theta_i$ is $1$-periodic and reach a minimum at $x_i\in\mathbb{R}$. Let define the map $\widetilde{\theta_2}\colon\mathbb{R}\rightarrow\mathbb{R}$ by:
$$\overline{\theta_2}(x):=\theta_2(x+x_1-x_2).$$
Notice that $\overline{\theta_2}$ reach a minimum at $x_1$. Now, for $t\in[0,1]$ let define the continuous map $\alpha_t\colon\mathbb{R}\rightarrow\mathbb{R}$ by:
$$\alpha_t(x)=(1-t)\theta_1(x)+t\overline{\theta_2}(x).$$
Notice that $\alpha_t$ is non-constant, indeed otherwise for all $x\in\mathbb{R}$, one has:
$$(1-t)\underbrace{(\theta_1(x)-\theta_1(x_1))}_{\geqslant 0}+t\underbrace{(\overline{\theta_2}(x)-\overline{\theta_2}(x_2))}_{\geqslant 0}=0.$$
But since $t$ and $1-t$ are both positive and at least one of them is nonzero, $\theta_1$ or $\overline{\theta_2}$ would have to be constant. Notice that $\theta_2$ is constant if and only if $\overline{\theta_2}$ is. Therefore, there exists $i\in\{1,2\}$ such that $\theta_i$ is constant. In particular, ${g_i}'$ is constant and nonzero and $g_i(0)$ must be distinct from $g_i(1)$, which is a contradiction. Finally, let define the continuous map $A\colon\mathbb{S}^1\times[0,1]\rightarrow\mathbb{S}^1$ by: 
$$A(x,t)=p\left(\alpha(x)\right).$$
One has $A(\cdot,0)={g_1}'$ and $A(\cdot,1)=p\circ\widetilde{\theta_2}$. Furthermore, for all $t\in[0,1]$, $A$ is non-constant when restricted to $\mathbb{S}^1\times\{t\}$. Otherwise, there would exist $t\in[0,1]$ such that the continuous map $\alpha_t-\alpha_t(0)$ takes its values in $\mathbb{Z}$ and would have been be constant, which is not the case since $\alpha_t$ ins non-constant. To conclude, it suffices to see that $p\circ\widetilde{\theta_2}$ and ${g_2}'$ are homotopic, which can be accomplished through: $$B(x,t)=p(\theta_2(x+t(x_1-x_2))).$$
Notice that for all $t\in[0,1]$, $B(\cdot,t)$ is also non-constant (same argument). Therefore, the concatenation of $A$ and $B$ leads to a suitable $H$.
