Well, turns out my mistake was not so simple after all: The map is NOT Frechet differentiable. Consider $g=0$, and $v_p(t)=\sqrt{1-p}t^{-p/2}=a_pt^{-p/2}$ so that $\| v_p\|_{L^2[0,1]}=1$ for all $0<p<1$ and the quotient we're looking at is
$$
\dfrac{\| \sin(hv_p)-hv_p\|_{L^2[0,1]}}{|h|} = \dfrac{1}{|h|} \left(\int_0^1 (\sin ha_pt^{-p/2})-(ha_pt^{-p/2}))^2\, dt \right)^{1/2}.
$$
Now let $b_p= \left(\frac{a_p h}{100}\right)^{2/p} <1$, and notice that for $0<t<b_p$ we have
$$
|\sin ha_pt^{-p/2}-(ha_pt^{-p/2})| \geq ha_p t^{-p/2}-1 \geq \dfrac{ha_p}{2t^{p/2}},
$$
and so
$$
\dfrac{1}{|h|} \left(\int_0^1 (\sin ha_pt^{-p/2})-(ha_pt^{-p/2}))^2\, dt \right)^{1/2} \geq \dfrac{1}{|h|} \left( \int_0^{b_p} \left(\dfrac{ha_p}{2t^{p/2}}\right)^2\, dt \right)^{1/2}= \dfrac{a_p}{2} \left( \int_0^{b_p} \dfrac{dt}{t^{p}}\, dt \right)^{1/2}= \dfrac{a_p}{2\sqrt{1-p}} b_p^{(1-p)/2}= \dfrac{1}{2}\left( \dfrac{a_ph}{100}\right)^{(1-p)/p}.
$$
To finish the argument, choose $h=1/n$ and $p_n= \frac{n}{n+1}$, to see that the difference quotient tends to $1/2$.
I hope I didn't miss anything, be sure to double check!
Another example that may help clarify the distinction is the following. Let $C = \{(x_1,x_2): 0< x_1^3<x_2<x_1^2\}$ and let $f = \mathbb 1_C$ be the indicator function of $C$, thus $f(x_1,x_2)=1$ if $(x_1,x_2) \in C$ and $f(x_1,x_2)=0$ otherwise.
Since $\mathbb R^2$ is finite-dimensional, all linear maps are continuous, so $f$ is Gateaux differentiable at $a \in \mathbb R^2$ precisely when the directional derivatives of $f$ at $a$ are given by a linear map.
We claim that the directional derivatives of $f$ at $0_2 \in \mathbb R^2$, all exist and indeed they all vanish: if $v = (v_1,v_2) \in \mathbb R^2$ then if $v_1.v_2\leq 0$ then $f(t.v)=0$ for all $t$ and so $\partial_v f(0)=0$. If $v_1v_2>0$ then by symmetry we may assume that $v_1,v_2$ are positive, so that $f(t.v)=0$ when $t<0$ and if $t>0$ then $f(tv_1,tv_2)=1$ if $t^3v_1^3<tv_2<t^2v_1^2$, or $v_2/v_1^2<t<\sqrt{v_2/v_1^3}$, so that $f(t.v)=0$ for $t<v_2/v_1^2$. Hence for $|t|<v_2/v_1^2$ we have $f(t.v)=0$ so that $\partial_v f(0)=0$ for $v$ with $v_1v_2>0$ also.
Thus, once we fix a direction $\mathbb R.v$, that is, restricting $f$ to any line $\mathbb R.v$, the function $f$ is identically $0$ near $0_2\in\mathbb R^2$. It is immediate that $Df(0_2) =0_{\text{Mat}_2(\mathbb R)}$, i.e. the Gateaux derivative exists, but $f$ is obviously not even continuous at $0_2$.
What this example is emphasizing is that, once $\dim(V)>1$, the requirement that a limit exists as $h\to 0$, i.e. as $\|h\|\to 0$ is strictly stronger than the requirement that the limits exist as you tend to zero along each ray through the origin, even where all these directional limits are compatible in the sense that their values are a linear function of the direction: in the case of $C$, if $p(t) = (t,1/2(t^2+t^3))$, then $f(p(t))=1$ for all $t \in (0,1)$ and $p(t)\to 0_2$ as $t\to 0$.
Addendum: If you want an example where $f$ is continuous, and $Df(0_2)$ exists but $f$ is not Frechet differentiable at $0_2$, then let $f(x) = \|x\|\mathbb 1_C$. Then, since $\|f(x)\|\leq \|x\|$ for all $x$, $f$ is clearly continuous at the origin. The same reasoning as above shows that the directional derivatives $\partial_v f(0_2)$ all exist and are zero. To see that $f$ is nevertheless not Frechet differentiable at $x=0_2$, note first that, if the Frechet derivative exists, it must be equal to the Gateaux derivative, and since this vanishes in our case, the condition that $f$ is Frechet differentiable at $0_2$ becomes $|f(h)|/\|h\| \to 0$ as $h \to 0$. Since $|f(p(t))|/\|p(t)\| = 1$ for all $t \in (0,1/2)$, it follows that the Frechet derivative does not exist at $0_2$.
Best Answer
$F(u+h)-F(u) \to 0$ as $h \to 0$ because $Ah \to 0$ as $h \to 0 $ and $r(u,h) =\|h\|_U \frac {r(u,h)} {\|h\|_U} \to 0$ too. [$\|Ah\| \leq \|A\| \|h\|_U \to 0$].