Much much simpler, no need for Lefschtz.
Consider a lift $F$. Notice that $F(0)-0=(F(1)-1)+2$ (make the computations, it is orientation reversing). So, the $2$ gives you the answer!
Presumably you mean that $f^2$ has zero rotation number (makes no sense otherwise according to the definitions, in particular in KH). But indeed it follows as you say.
For $\phi = f, g$ we set $\phi^0 = id$ and $\phi^{-m} = (\phi^{-1})^m$ for $m \in \mathbb N$. This defines $\phi^n$ for all $n \in \mathbb Z$. We have $\phi^n(p) \in (a,b)$ for all $n$ and $\phi^{n+1}(p) = \phi(\phi^n(p)) > \phi^n(p)$ for all $n \in \mathbb Z$. We claim that $\phi^n(p) \to b$ as $n \to +\infty$, $\phi^n(p) \to a$ as $n \to -\infty$.
Clearly $\lim_{n \to +\infty} \phi^n(p) = b' \le b$. Note that all $\phi^n(p) < b'$. Assume $b' < b$. Then $b' \in (a,b)$ and $\phi(b') > b'$, i.e. $\phi(b') - b' > 0$. By continuity there exists $\epsilon > 0$ such that $\phi(t) - t > 0$ for all $t \in (a,b)$ with $\lvert t - b' \rvert < \epsilon$. But $\lvert \phi^n(p) - b' \rvert < \epsilon$ for sufficiently large $n$, thus $\phi^n(p) > b'$ which is a contradiction. The case $n \to -\infty$ is treated similarly.
- Each $h_n$ is a homeomorphism.
Proof. Since $\phi^n$ is a homeomorphism, the restriction $\overline{\phi^n} : [p,\phi(p)] \stackrel{\phi^n}{\to} [\phi^n(p),\phi^{n+1}(p)]$ is a homeomorphism (the notation $\overline{\phi^n}$ is used as a distinction to $\phi^n$). Note that $(\overline{\phi^n})^{-1}$ is the restriction of $\phi^{-n}$ to $[\phi^n(p),\phi^{n+1}(p)]$.
Thus $h_n = \overline{g^n} \circ h_0 \circ (\overline{f^n})^{-1}$ is a homeomorphism. This is what you wrote as $h_n = g^n \circ h_0 \circ f^{-n}$. In fact, for $t \in [f^n(p),f^{n+1}(p)]$ we have
$$h_n(t) = g^n(h_0(f^{-n}(t))) \tag{1} .$$
By construction
$$h_n(f^n(p)) = g^n(p), h_n(f^{n+1}(p)) = g^{n+1}(p) \tag{2} .$$
For $t \in [f^n(p),f^{n+1}(p)]$ we get
$$h_{n+1}(f(t)) = g^{n+1}(h_0(f^{-(n+1)}(f(t)))) = g^{n+1}(h_0(f^{-n}(t))) = g(g^n(h_0(f^{-n}(t)))) \\ = g(h_n(t)) \tag{3}.$$
- Define
$$h(t) = \begin{cases} a & t = a \\ h_n(t) & t \in [f^n(p),f^{n+1}(p)] \\ b & t = b\end{cases}$$
This is well-defined by $(2)$ because $[f^n(p),f^{n+1}(p)] \cap [f^{n+1}(p),f^{n+2}(p)] = \{f^{n+1}(p)\}$. It is also shows that $h$ is continuous on $\bigcup_{n \in \mathbb Z} [f^n(p),f^{n+1}(p)] =(a,b)$. Continuity in $a$ and $b$ follows from the fact that $\phi^n(p) \to b$ as $n \to +\infty$, $\phi^n(p) \to a$ as $n \to -\infty$.
By construction $h$ is a continuous bijection, thus a homeomorphism, such that $h \circ f = g \circ h$.
Best Answer
If $f(0)=0$, we are done.
If $f(0)>0$, then $f(f(0))<f(0)$, and if $f(0)<0$, then $f(f(0))>f(0)$. In both cases, we have poitns $x_1,x_2$ with $f(x_1)<x_1$ and $f(x_2)>x_2$. Then the Intermediate Value Theorem tells us that the continuous function $x\mapsto f(x)-x$ has a zero between $x_1$ and $x_2$, i.e., $f$ has a fixed point.