Given the space of bump functions $\mathcal{D}(\mathbb{R}^n)$, smooth and with compact support,
$$\mathcal{D}(\mathbb{R}^n) := \{ v \in C^\infty(\mathbb{R}^n) : \mathrm{supp}(v) \ \mathrm{is\ compact} \}$$
and the Schwartz space $\mathcal{S}(\mathbb{R})^n$, of functions smooth and "rapidly decreasing" toward $\infty$
$$\mathcal{S}(\mathbb{R}^n) := \{ v \in C^\infty(\mathbb{R}^n) : \alpha \in \mathbb{N}^r, r,m\in \mathbb{N} \implies D^\alpha v = o(|\mathbf{x}|^{-m}) \ \mathrm{as}\ |\mathbf{x}| \to \infty \}$$
where obviously
$$\mathcal{D}(\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n)$$
what are the relationship between the convergence $v_k \to v$ in $\mathcal{D}(\mathbb{R}^n)$ of a sequence $\{v_k\} \subset \mathcal{D}(\mathbb{R}^n)$
$$ \exists K\subset\mathbb{R}^n \ \mathrm{compact}: \forall k \in \mathbb{N}: \ \mathrm{supp}(v_k)\subset K $$ $$\forall \alpha \in \mathbb{N}^r, r\in \mathbb{N}: D^\alpha v_k \to D^\alpha v \ \mathrm{as} \ k\to\infty \ \mathrm{uniformly\ in}\ \mathbb{R}^n$$
and the convergence $v_k \to v$ in $\mathcal{S}(\mathbb{R}^n)$ of the same sequence $\{v_k\} \subset \mathcal{D}(\mathbb{R}^n)$
$$\forall \beta \in \mathbb{N}^n, \forall \alpha \in \mathbb{N}^r, r\in \mathbb{N}: \mathbf{x}^\beta D^\alpha v_k \to \mathbf{x}^\beta D^\alpha v \ \mathrm{as} \ k\to\infty \ \mathrm{uniformly\ in}\ \mathbb{R}^n$$ ?
When one implies the other? There are conditions under which the second implies the first?
EDIT. To be more explicit... If $v_k \to v$ in $\mathcal{D}(\mathbb{R}^n)$, because $\mathcal{D}(\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n)$, one can ask himself if $v_k \to v$ also in $\mathcal{S}(\mathbb{R}^n)$, and it's someway obvious the answer is positive.
If $v_k \to v$ in $\mathcal{S}(\mathbb{R}^n)$, to ask himself if $v_k \to v$ also in $\mathcal{D}(\mathbb{R}^n)$, at very minimum we need to suppose all $v_k$ are also in $\mathcal{D}(\mathbb{R}^n)$. Is this enough to prove also $v$ is in $\mathcal{D}(\mathbb{R}^n)$ and $v_k \to v$ also in $\mathcal{D}(\mathbb{R}^n)$? No, because for every $v \in \mathcal{S}(\mathbb{R}^n)$ there is a sequence ${v_k} \subset \mathcal{D}(\mathbb{R}^n)$ such that $v_k \to v$ in $\mathcal{S}(\mathbb{R}^n)$. And if all $v_k$ and $v$ are in $\mathcal{D}(\mathbb{R}^n)$ and $v_k \to v$ in $\mathcal{S}(\mathbb{R}^n)$ is this enough? If not what else should I suppose?
Obviously, the first convergence implies the second. If the supports of all $v_k$ are in the same compact set and uniform convergence on this set is given, the convergence in $\mathcal{S}$ follows immidiately as the polynomials are bounded on this set.
The second convergence does not imply the first in general, an example are smoothly cut-off $e^{-x^2}$, i.e. something like $f_n(x)=e^{-x^2}$ for $|x|<n$ and $f_n(x)=0$ for $|x|>n+1$ and smooth in between.
EDIT after the edit in the question:
No, you may imagine the sequence $g_n= \frac{1}{n} f_n$ with $f_n$ as above, then $(g_n)_n\to 0$ in $\mathcal{S}$ but not in $\mathcal{D}$ even though all functions and the limit function are elements of $\mathcal{D}$. The crucial property for the convergence in $\mathcal{D}$ is the uniformly bounded support of the functions.