Let $(X,\left\|\;\cdot\;\right\|)$ be a Banach space and $A:X\to\mathbb R$ be Fréchet differentiable, i.e. $\exists{\rm D}A:X\to\mathfrak L(X,\mathbb R)$$^1$ with $$\lim_{\left\|h\right\|\to 0}\frac{\left|A(x_0+h)-A(x_0)-{\rm D}A(x_0)h\right|}{\left\|h\right\|}=0\;\;\;\text{for all }x_0\in X\;.$$
Let $$B(t):=x_0+th\;\;\;\text{for }t\in\mathbb R$$ for some $x_0,h\in X$ and $$f(t):=(A\circ B)(t)\;\;\;\text{for }t\in\mathbb R\;.$$ It's easy to show that ${\rm D}B:\mathbb R\to\mathfrak L(\mathbb R,X)$ is given by $${\rm D}B(t)=U\;\;\;\text{for all }t\in\mathbb R\;,$$ where $$Ut:=th\;\;\;\text{for }t\in \mathbb R\;$$
Thus, we can conclude that $${\rm D}f(t)={\rm D}(A\circ B)(t)={\rm D}A\left(B(t)\right)\circ{\rm D}B(t)={\rm D}A\left(B(t)\right)\circ U\in\mathfrak L(\mathbb R,X)$$ for all $t\in\mathbb R$.
How can I compute the second Fréchet derivative ${\rm D^2}f$?
$^1$ Let $\mathfrak L(U,V)$ be the space of bounded, linear operators from $U$ to $V$.
We start from $$ Df(t) = DA\bigl(B(t)\bigr)U $$ By the product rule, we have (note that the product rule holds for composition of linear maps, as the product is bilinear and continuous): $$ D^2 f(t) = D\Bigl(DA\bigl(B(t)\bigr)\Bigr)U + DA\bigl(B(t)\bigr)DU $$ As $U$ is a constant, and does not depend on $t$, $DU = 0$, we have, by the chain rule $$ D^2 f(t) = D^2 A\bigl(B(t)\bigr)[U, U] $$ (as $DB(t) = U$). Here $[U, U]$ denotes the map $\mathbf R^2 \to X^2$ given by $[U, U](s,t) =[U(s), U(t)]$, so the bilinear map $D^2 f(t) \colon \mathbf R^2 \to \mathbf R$ is given by $$ D^2 f(t)[s_1, s_2] = D^2 A\bigl(B(t)\bigr)[Us_1, Us_2] = D^2 A(x_0 + th)[s_1h, s_2 h] $$