[proofplan]
Write the maps componentwise and prove the result by computing first and second coordinate partial derivatives. The first derivative formula is the ordinary coordinate chain rule. Differentiating that formula once more gives an explicit expression for every second [partial derivative](/page/Partial%20Derivative) of each component of $G\circ F$. Each term in that expression is a finite product and sum of continuous functions, so all first and second partial derivatives are continuous on $U$.
[/proofplan]
[step:Write the composition componentwise]
For $1\le p\le m$, let
\begin{align*}
F_p:U&\to\mathbb R
\end{align*}
denote the $p$-th component of $F$. For $1\le a\le k$, let
\begin{align*}
G_a:V&\to\mathbb R
\end{align*}
denote the $a$-th component of $G$. Since $F\in C^2(U;\mathbb R^m)$, each $F_p$ belongs to $C^2(U)$. Since $G\in C^2(V;\mathbb R^k)$, each $G_a$ belongs to $C^2(V)$.
Define, for each $1\le a\le k$, the scalar component
\begin{align*}
H_a:U&\to\mathbb R
\end{align*}
\begin{align*}
x&\mapsto G_a(F(x)).
\end{align*}
Then $G\circ F=(H_1,\dots,H_k)$. Hence it is enough to prove that every $H_a$ belongs to $C^2(U)$.
[/step]
[step:Compute the first partial derivatives of each component]
Fix $1\le a\le k$ and $1\le i\le n$. Since $F(U)\subset V$, the expression $G_a(F(x))$ is defined for every $x\in U$. Applying the coordinate chain rule to the scalar map $G_a\circ F:U\to\mathbb R$ gives, for every $x\in U$,
\begin{align*}
\partial_{x_i}H_a(x)=\sum_{p=1}^{m}\partial_{y_p}G_a(F(x))\,\partial_{x_i}F_p(x),
\end{align*}
where $y=(y_1,\dots,y_m)$ denotes the standard coordinate variable on $\mathbb R^m$.
For each $p$, the function $x\mapsto \partial_{y_p}G_a(F(x))$ is continuous on $U$ because $\partial_{y_p}G_a:V\to\mathbb R$ is continuous and $F:U\to V$ is continuous. Also $x\mapsto\partial_{x_i}F_p(x)$ is continuous on $U$ because $F_p\in C^2(U)$. Therefore each product in the finite sum is continuous, and $\partial_{x_i}H_a:U\to\mathbb R$ is continuous.
[/step]
[step:Differentiate the first derivative formula once more]
Fix $1\le a\le k$ and $1\le i,j\le n$. We differentiate the formula for $\partial_{x_i}H_a$ with respect to $x_j$. For each $1\le p\le m$, the product rule gives
\begin{align*}
\partial_{x_j}\left(\partial_{y_p}G_a(F(x))\,\partial_{x_i}F_p(x)\right)
=
\partial_{x_j}\left(\partial_{y_p}G_a(F(x))\right)\partial_{x_i}F_p(x)
+
\partial_{y_p}G_a(F(x))\,\partial_{x_j}\partial_{x_i}F_p(x).
\end{align*}
Applying the coordinate chain rule to the scalar map
\begin{align*}
\partial_{y_p}G_a\circ F:U&\to\mathbb R
\end{align*}
gives
\begin{align*}
\partial_{x_j}\left(\partial_{y_p}G_a(F(x))\right)
=
\sum_{q=1}^{m}\partial_{y_q}\partial_{y_p}G_a(F(x))\,\partial_{x_j}F_q(x).
\end{align*}
Substituting this into the product-rule expression and summing over $p$ yields
\begin{align*}
\partial_{x_j}\partial_{x_i}H_a(x)
=
\sum_{p=1}^{m}\sum_{q=1}^{m}\partial_{y_q}\partial_{y_p}G_a(F(x))\,\partial_{x_j}F_q(x)\,\partial_{x_i}F_p(x)
+
\sum_{p=1}^{m}\partial_{y_p}G_a(F(x))\,\partial_{x_j}\partial_{x_i}F_p(x).
\end{align*}
[guided]
We already know the first derivative of $H_a=G_a\circ F$:
\begin{align*}
\partial_{x_i}H_a(x)=\sum_{p=1}^{m}\partial_{y_p}G_a(F(x))\,\partial_{x_i}F_p(x).
\end{align*}
To prove $H_a\in C^2(U)$, we must differentiate this expression once more and then prove the resulting second partial derivative is continuous.
Fix indices $1\le i,j\le n$. The summation is finite, so we may differentiate term by term. For a fixed $p$, the $p$-th summand is a product of the two scalar functions
\begin{align*}
x&\mapsto \partial_{y_p}G_a(F(x)),
\end{align*}
and
\begin{align*}
x&\mapsto \partial_{x_i}F_p(x).
\end{align*}
The product rule gives
\begin{align*}
\partial_{x_j}\left(\partial_{y_p}G_a(F(x))\,\partial_{x_i}F_p(x)\right)
=
\partial_{x_j}\left(\partial_{y_p}G_a(F(x))\right)\partial_{x_i}F_p(x)
+
\partial_{y_p}G_a(F(x))\,\partial_{x_j}\partial_{x_i}F_p(x).
\end{align*}
It remains to compute the derivative of $x\mapsto\partial_{y_p}G_a(F(x))$. Since $G_a\in C^2(V)$, the partial derivative $\partial_{y_p}G_a:V\to\mathbb R$ is $C^1$. Applying the coordinate chain rule to $\partial_{y_p}G_a\circ F$ gives
\begin{align*}
\partial_{x_j}\left(\partial_{y_p}G_a(F(x))\right)
=
\sum_{q=1}^{m}\partial_{y_q}\partial_{y_p}G_a(F(x))\,\partial_{x_j}F_q(x).
\end{align*}
Substituting this expression into the product-rule formula and summing over $p$ gives the second derivative formula
\begin{align*}
\partial_{x_j}\partial_{x_i}H_a(x)
=
\sum_{p=1}^{m}\sum_{q=1}^{m}\partial_{y_q}\partial_{y_p}G_a(F(x))\,\partial_{x_j}F_q(x)\,\partial_{x_i}F_p(x)
+
\sum_{p=1}^{m}\partial_{y_p}G_a(F(x))\,\partial_{x_j}\partial_{x_i}F_p(x).
\end{align*}
This formula is the key point: it expresses every second partial derivative of the composition using only first and second partial derivatives of $F$ and $G$.
[/guided]
[/step]
[step:Show the second partial derivatives are continuous]
Fix $1\le a\le k$ and $1\le i,j\le n$. In the formula
\begin{align*}
\partial_{x_j}\partial_{x_i}H_a(x)
=
\sum_{p=1}^{m}\sum_{q=1}^{m}\partial_{y_q}\partial_{y_p}G_a(F(x))\,\partial_{x_j}F_q(x)\,\partial_{x_i}F_p(x)
+
\sum_{p=1}^{m}\partial_{y_p}G_a(F(x))\,\partial_{x_j}\partial_{x_i}F_p(x),
\end{align*}
each factor is continuous on $U$. Indeed, $F:U\to V$ is continuous, each $\partial_{y_q}\partial_{y_p}G_a:V\to\mathbb R$ and each $\partial_{y_p}G_a:V\to\mathbb R$ is continuous because $G_a\in C^2(V)$, and each $\partial_{x_j}F_q$, $\partial_{x_i}F_p$, and $\partial_{x_j}\partial_{x_i}F_p$ is continuous because $F_p\in C^2(U)$.
Finite products and finite sums of continuous real-valued functions are continuous. Therefore $\partial_{x_j}\partial_{x_i}H_a:U\to\mathbb R$ is continuous for every $1\le i,j\le n$.
[/step]
[step:Conclude that the composition is $C^2$]
For each $1\le a\le k$, the component $H_a:U\to\mathbb R$ has continuous first partial derivatives and continuous second partial derivatives on $U$. Hence $H_a\in C^2(U)$. Since
\begin{align*}
G\circ F=(H_1,\dots,H_k),
\end{align*}
all components of $G\circ F$ belong to $C^2(U)$. Therefore
\begin{align*}
G\circ F\in C^2(U;\mathbb R^k).
\end{align*}
This is the desired conclusion.
[/step]
