Everything that follows is from Rudin's Functional Analysis: $\def\L{\Lambda} \def\DDD{\mathcal{D}} \def\sbe{\subseteq} \def\W{\Omega} \def\RR{\mathbb{R}} \def\CC{\mathbb{C}} $
Below $\DDD$ is the space of test (smooth, compactly supported) functions from an open set $\Omega\sbe\RR^n$ to $\CC$. Such space is given a complete, unmetrizable topology $\tau$.
Similarly $\DDD_K$ is the space of test functions $\Omega\to\CC$ whose support lies in the compact set $K$. Each $\DDD_K$ has a Fréchet space topology $\tau_K$ which corresponds with the subspace topology under $\tau$.
Theorem 6.6: suppose $\L$ is a linear mapping of $\DDD$ into a lctvs $Y$. Then the following are equivalent:
a) $\L$ is continuous.
b) $\L$ is bounded.
c) If $\phi_i\to 0$ in $\DDD(\W)$, then $\L\phi_i\to0$ in $Y$.
d) The restriction of $\L$ to any $\DDD_K\sbe\DDD(\W)$ is continuous.
The proof shows a)$\implies$b)$\implies$c)$\implies$d)$\implies$a). I struggle to understand the steps b)$\implies$c) and d)$\implies$a):
b)$\implies$c): there is a compact subset $K\sbe\W$ that contains the support of every $\phi_i$, thus the $\phi_i$ are members of the subset $\DDD_K$ and $\phi_i\to 0$ holds in $\DDD_K$. The restriction of $\L$ to this $\DDD_K$ is bounded (why?) and since $\DDD_K$ is metrizable it follows that $\L\phi_i\to 0$ in $Y$.
d)$\implies$a): let $U$ be a convex balanced neighborhood of $0$ in $Y$, and put $V=\L^{-1}(U)$. Then $V$ is convex and balanced (why?). Now $V$ is open if and only if $\DDD_K\cap V\in\tau_K$ for every compact $K\sbe \W$ (I understand the only if implication only). This proves the equivalence of a) and d) (how?).
Let start with $(b) \to (c)$
As $\mathcal{D}_K$ is the subspace topology of $\mathcal{D}$ if $E$ is a bounded set in $\mathcal{D}_K$ than it is bounded in $\mathcal{D}$.
A linear map $\Lambda : \mathcal{D} \to Y$ between tvs is bounded if, for any $E$ bounded $\Lambda(E)$ is bounded in $Y$.
Assume $\Lambda |_{\mathcal{D}_K}$ is not bounded, than there is $E$ bounded in $\mathcal{D}_K$
such that $\Lambda(E)$ is not bounded, but $E$ is bounded in $\mathcal{D}$, so this contraddict condition $(b)$ and $\Lambda_{\mathcal{D}_K}$ is bounded
Now for $(d) \to (a)$.
The antimage of a linear map of a balanced set is balanced. Indeed, for any $a$ such that $|a|=1$ we have $$ a \Lambda^{-1}(U)=\Lambda^{-1} (aU) \subseteq \Lambda^{-1}(U) $$ The proof that antiimage of convex sets is convex is very similar. So $V$ is convex and balanced.
By theorem $6.5$ $(a)$ of the second edition of Rudin functional analysis you have that $V$ is open iif $V \cap D_K \in \tau_K$. So you have $\Lambda^{-1}(U)$ is open in $\mathcal{D}$ iif $\Lambda^{-1}(D) \cap D_K$ is open in $\tau_K$
The first condition is the definition of continuity in $\mathcal{D}$ (i.e condition $(a)$).
The second condition is the definition of continuity in $\mathcal{D}_K$ (i.e condition $(d)$)
So they are equivalent