Let $T>0$, $I:=(0,T)$ and $V$ be a normed $\mathbb R$-vector space. We can easily show$^1$ that $$P(I,V):=\operatorname{span}\left\{\tau\varphi:(\tau,\varphi)\in C^\infty(I)\times V\right\}$$ is a dense subset of $C_c^\infty(I,V)$ with respect to the supremum norm.
Now let $k\in\mathbb N_0$ and $C_b^k(I,V)$ denote the space of bounded $k$-times continuously differentiable functions from $I$ to $V$ equipped with $$\left\|u\right\|_{C_b^k(I,\:V)}:=\max_{0\le i\le k}sup_{t\in I}\left\|u^{(i)}(t)\right\|_V.$$ Are we able to show that $P(I,V)$ is a dense subset of $C_c^k(I,V)$?
Let $u\in C_c^k(I,V)$ and $\varepsilon>0$. By the aforementioned result, there are $v_0,\ldots,v_k\in P(I,V)$ s.t. $$\max_{0\le i\le k}\left\|u^{(i)}-v_i\right\|_{C_b(I,\:V)}<\varepsilon\tag7.$$ The claim would follow, if we could show or pick $v_0$ s.t. $v_0^{(i)}=v_i$ for all $i\in\{0,\ldots,k\}$.
$^1$ Let $u\in C_c^\infty(I,V)$ and $\varepsilon>0$. Since $u$ is uniformly continuous, there is a $\delta>0$ s.t. $$|s-t|<\delta\Rightarrow\left\|u(s)-u(t)\right\|_V<\varepsilon\tag1$$ for all $s,t\in I$. Since $K:=\operatorname{supp}u$ is a compact and hence totally bounded subset of $I$, there is a finite $F\subseteq K$ s.t. $$K\subseteq\bigcup_{t\in F}\left\{s\in I:|s-t|<\delta\right\}\tag2$$ which in turn implies that there are $k\in\mathbb N$ and $t\in I^k$ with $$0:<t_{i+1}-t_i<\delta\tag3\;\;\;\text{for all }i\in\{1,\ldots,k-1\}$$ and $$(0,t_1]\cup[t_k,T)\subseteq I\setminus K\tag4.$$ Let $t_0:=0$ and $t_{k+1}:=T$. Then $((t_{i-1},t_{i+1})_{1\le i\le k}$ is an open cover of $I$ and hence there is a $C^\infty$-partition of unity $(\tau_i)_{1\le i\le k}$ subordinate to $((t_{i-1},t_{i+1})_{1\le i\le k}$. Let $$v(t):=\sum_{i=1}^k\tau_i(t)u(t_i)\;\;\;\text{for }t\in I.$$ By definition, $\sum_{i=1}^k\tau_i=1$ and $\tau_1,\ldots,\tau_k\ge0$ and hence $$\left\|u(t)-v(t)\right\|_V\le\sum_{i=1}^k\tau_i(t)\underbrace{\left\|u(t)-u(t_i)\right\|_V}_{<\:\varepsilon}<\varepsilon\tag5$$ by $(1)$; i.e. $$\left\|u-v\right\|_{C_b(I,V)}<\varepsilon\tag6.$$