= Solution
The <contraction mapping theorem> states that a <contraction mapping> $T:X\to X$ on a nonempty <complete metric space> has a unique <fixed point>. Here contraction means that $d(Tx,Ty)\leq qd(x,y)$ for all $x,y$, with one constant $0\leq q<1$. Choose $x_0\in X$ and define $x_{n+1}=Tx_n$. Successive distances satisfy $d(x_{n+1},x_n)\leq q^nd(x_1,x_0)$, so for every $m\geq1$,
$$
d(x_{n+m},x_n)\leq\sum_{j=n}^{n+m-1}q^jd(x_1,x_0)\leq\frac{q^n}{1-q}d(x_1,x_0).
$$
Thus $(x_n)$ is a <Cauchy sequence>. Completeness gives $x_n\to x_*$. Since a <contraction mapping> is a <continuous function>, $Tx_* =\lim Tx_n=\lim x_{n+1}=x_*$. If $y_*$ were another <fixed point>, $d(x_*,y_*)=d(Tx_*,Ty_*)\leq qd(x_*,y_*)$, forcing distance zero. This proves both existence and uniqueness.
Now suppose $T=f^k$ is a <contraction mapping>, where $k$ is a positive integer. Let $y$ be its unique <fixed point>. Since $f$ commutes with its own iterate, $T(fy)=f(Ty)=fy$. Thus $fy$ is also a <fixed point> of $T$, so $fy=y$. Any <fixed point> of $f$ is a <fixed point> of $T$, and must equal $y$. \b[A contractive iterate therefore gives a unique <fixed point> of $f$], without assuming continuity of $f$ itself.
The space $X=C([0,1],\mathbb R)$ is complete for the <uniform norm>. Indeed, a uniformly <Cauchy sequence> converges pointwise by completeness of $\mathbb R$; passing to the limit in its uniform Cauchy bound shows <uniform convergence>, and the uniform limit of <continuous functions> is a <continuous function>. The operator $F$ maps this space to itself: $s\mapsto\phi(h(s),s)$ is a <continuous function>, and its integral plus the continuous $g$ is a <continuous function>.
For $n=0$ the desired bound is the definition of $\|h-k\|_\infty$. Suppose it holds for $n$. The <Lipschitz condition> and the fact that the same $g$ cancels give
$$
\begin{aligned}|F^{n+1}(h)(t)-F^{n+1}(k)(t)|&\leq\int_0^t M|F^n(h)(s)-F^n(k)(s)|\,ds\\&\leq\frac{M^{n+1}}{n!}\|h-k\|_\infty\int_0^t s^n\,ds\\&=\frac{M^{n+1}t^{n+1}}{(n+1)!}\|h-k\|_\infty.
\end{aligned}
$$
This proves the estimate by induction. Taking the supremum over $0\leq t\leq1$, the Lipschitz constant of $F^n$ is at most $M^n/n!$. These constants tend to zero, since the ratio of successive terms is $M/(n+1)$. Some iterate is consequently a contraction. The preceding result gives a unique <fixed point> of $F$, which is exactly $\boxed{\text{the unique continuous solution of the integral equation on }[0,1]}$. No assumption $M<1$ is needed.
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