Proof by contradiction in Constructive Mathematics
One rule is negation-introduction; that is "If we derive a contradiction under an assumption, then we may infer that the premises entail the negation of the assumption." This is a "Proof of Negation". $$\begin{split}\Gamma, p&\vdash \bot\\\hline\Gamma&\vdash\lnot p\end{split}\tag {1, $\lnot$i}$$
Well, what if we derive a contradiction when we assume a negation? Applying this rule we would infer that the premises entail the negation of that negation; ie a 'double negation'. $$\begin{split}\Gamma, \lnot p&\vdash \bot\\\hline\Gamma&\vdash\lnot\lnot p\end{split}\tag 2$$
This is all intuitionistically valid. The difference between intuitionistic (contructive) and classical logic is the rule of double-negation-elimination (DNE), is only admissible under the later; it says "if we can derive a double negation of a proposition we may infer that we can derive the proposition." $$\begin{split}\Gamma &\vdash \lnot\lnot p\\\hline\Gamma&\vdash p\end{split}\tag {3, DNE}$$
Putting (2) and (3) together and we have "Proof by Reduction to Absurdity", and this is why it is not intuitionistically valid. $$\begin{split}\Gamma, \lnot p&\vdash \bot\\\hline\Gamma&\vdash p\end{split}\tag{4, RAA}$$
The second is how you prove negation.
When you are proving "not $A$", you are proving "if $A$, then contradiction", because this is literally how "not $A$" is defined in constructive logic.
When you are proving "$A$", it does not suffice to prove "if not $A$, then contradiction", because $A$ is not defined this way (that would be circular!) and there is no general rule that tells you that "if not $A$, then contradiction" implies $A$. There are situations when this does hold (for example, if $A$ is computable -- i.e., there exists a boolean $a \in \left\{\operatorname{True}, \operatorname{False}\right\}$ such that $A \Longleftrightarrow \left(a = \operatorname{True} \right)$), but these are theorems that have to be proven rather than consequences of a general axiom like TND. (For example, when $A$ is computable -- i.e., when you know a boolean $a \in \left\{\operatorname{True}, \operatorname{False}\right\}$ such that $A \Longleftrightarrow \left(a = \operatorname{True} \right)$ -- you can prove $A$ by proving "if not $A$, then contradiction" -- but the reason why this works is simply that you can distinguish cases based upon whether $a = \operatorname{True}$ or $a = \operatorname{False}$. So, on the level of axioms, you are not using TND, but rather emulating a "proof by contradiction" by inducting (= distinguishing cases) upon a boolean. If you don't know such a boolean $a$, then this trick does not work.)
Yes, these situations are dual in a sense, but "dual" does not mean "equally true" in mathematics. Even in classical mathematics. For example, a basis of a vector space may have any cardinality, but a basis of a dual of a vector space may not :)