Let me use the notation $Df$ and $D^2f$ to represent those maps and use $f'$ and $f''$ to mean the typical single variable notation for derivatives. Note that instead of thinking of $D^2f$ as a map $U \to \mathcal{L}(X,\mathcal{L}(X,Y))$, I think it is sometimes simpler to think of it as a map $U \to \mathcal{L}^2(X;Y)$ (the space of continuous bilinear maps $X\times X \to Y$; there is in fact a natural isometric (when equipped with operator norm) isomorphism between these spaces).
Then, of course, for all $x \in \Bbb{R}$, $f(x) = x^3$ implies $f'(x) = 3x^2, f''(x) = 6x, f'''(x), f^{(4)}(x) = 0$. In terms of the Frechet derivatives, these are:
\begin{align}
\begin{cases}
Df:\Bbb{R} \to \mathcal{L}(\Bbb{R}; \Bbb{R}),\quad Df_x[h]&= 3x^2h\\
D^2f:\Bbb{R} \to \mathcal{L}^2(\Bbb{R}; \Bbb{R}),\quad D^2f_x[h_1, h_2]&= 6x h_1 h_2\\
D^3f:\Bbb{R} \to \mathcal{L}^3(\Bbb{R}; \Bbb{R}),\quad Df_x[h_1, h_2, h_3]&= 6h_1h_2h_3
\end{cases}
\end{align}
for $k \geq 4$, $D^4f = 0$ is identically $0$. From these expressions it is clear that for every $k \in \Bbb{N}$, and every $x \in \Bbb{R}$, $D^kf_x$ is a symmetric continuous multilinear mapping $\underbrace{\Bbb{R} \times \dots\times \Bbb{R}}_{\text{$k$ times}} \to \Bbb{R}$.
Notice that $f'(x) = Df_x[1]$ and $f''(x) = D^2f_x[1,1]$. In general for any sufficiently differentiable function $f: \Bbb{R} \to \Bbb{R}$, we will have $f^{(k)}(x) = D^kf_x[\underbrace{1, \dots, 1}_{\text{$k$ times}}]$, or equivalently,
\begin{align}
D^kf_x[h_1, \dots, h_k] &= f^{(k)}(x) \cdot h_1 \cdots h_k.
\end{align}
(Try a proof by induction if you want to be super rigorous).
Edit: Responding to OP's comment
Let's prove a slightly more general theorem. If we have a function $f:\Bbb{R} \to Y$, we shall use the notation $f'(x)$ to mean the limit $\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$. Since the domain is $\Bbb{R}$ and the target space is a normed vector space $Y$, it makes sense to talk about this limit. I shall take for granted that you know the following fact:
For $x \in \Bbb{R}$, $f'(x)$ exists if and only if $Df_x$ exists in which case, $Df_x[h] = f'(x) \cdot h$.
This shouldn't be too hard to prove; if I remember correctly, its only a 3-4 line proof. Now, we can state the conclusion of the theorem slightly differently. Define the map $T: Y \to \mathcal{L}(\Bbb{R}, Y)$ as
\begin{align}
T(\alpha) &:= (h\mapsto h\cdot \alpha)
\end{align}
This can easily be checked to be a linear isometric isomorphism from $Y$ onto $\mathcal{L}(\Bbb{R},Y)$ (equipped with operator norm). With this notation, we can state the conclusion of the theorem as:
$f'(x)$ exists if and only if $Df_x$ exists in which case, $Df_x = T(f'(x))$, or equivalently, $Df = T \circ f'$.
Now, how do we calculate $D^2f_x$? We simply use the chain rule and the fact that derivatives of linear transformations are themselves:
\begin{align}
D^2f_x &= DT_{f'(x)} \circ D(f')_x \\
&= T \circ D(f')_x \tag{$T$ is linear} \\
&= T \circ [T(f''(x))],
\end{align}
where in the last line I applied the highlighted result to $f'$. In this notation, $D^2f_x$ is an element of $\mathcal{L}(\Bbb{R}, \mathcal{L}(\Bbb{R},Y))$. Now evaluate first on $h_1$ then evaluate on $h_2$, and then you'll see that by how $T$ is defined,
\begin{align}
(D^2f_x[h_1])[h_2] &= f''(x) \cdot h_1 h_2.
\end{align}
Or if we abuse notation slightly and refer to $D^2f_x$ as the associated bilinear continuous map, then
\begin{align}
D^2f_x[h_1, h_2] &= f''(x) \cdot h_1 h_2.
\end{align}
I leave it to you to prove inductively the following theorem:
For a function $f: U \subset \Bbb{R} \to Y$ ($U$ an open set), and for any integer $k\geq 0$, and any $x \in U$, the "usual derivative" $f^{(k)}(x)$ exists if and only if $D^kf_x$ exists (doesn't matter if you think of this as multilinear or not, because these two interpretations are related simply by application of a linear isomorphism... which doesn't affect differentiability). In this case,
\begin{align}
D^kf_x[h_1, \dots, h_k] &= f^{(k)}(x) \cdot h_1 \cdots h_k.
\end{align}
