Proof that $G$ is a yes-instance of IDS $\iff$ $f(G)$ is a yes-instance of SAT Consider the Independent Dominating Set problem with a directed graph $G=(V,E)$ as instance and the properties that:


$\forall (u,v) \in E, \{u,v\}\nsubseteq S$
$\forall v \in V: v \in S \lor \exists (u,v) \in E: u \in S$


then consider the function $f$ which is a many-one reduction from IDS to SAT for $G$:
$$f(G)= \bigwedge_{(u,v) \in E}(\neg x_u \lor \neg x_v) \land \bigwedge_{v \in V}(x_v \lor \bigvee_{(u,v) \in E}x_u)$$

Theorem: $G$ is a yes-instance of IDS $\iff$ $f(G)$ is a yes-instance of SAT

Proof:

"$\Leftarrow$": Assume that $f(G)$ is a yes-instance of SAT:

Consider the propositional atoms $x_u$ and $x_v$ which represent the vertices of an edge. $x$ is $true$ iff is is in $S$. To proof a CNF-formula true we have to proof all conjuncted subformulas true.

Now consider since $\bigwedge_{(u,v) \in E}(\neg x_u \lor \neg x_v)$ holds and tells us that only one of the vertices $u,v$ is allowed to be in $S$ at the same time, this implies $\forall (u,v) \in E, \{u,v\}\nsubseteq S$.

For the second part we know that $\bigwedge_{v \in V}(x_v \lor \bigvee_{(u,v) \in E}x_u)$ only evaluates to true if either $v \in S$ or any $u \in S$ which is connected by an edge to $v$. This implies $\forall v\in V: v \in S \lor \exists(u,v) \in E: u \in S$.

Therefore we are done.

Is my proof for "$\Leftarrow$" of the theorem correct?

Assumption: The proof is correct and the reduction is valid:

Now we know that IDS is NP-hard, for completeness of IDS we still have to show NP-membership of IDS, right? Furthermore we have to provide an efficient algorithm which uses non-determinism to show that IDS is a member of NP, correct?
Yes, your proof of $\Rightarrow$ is correct, but note that you still have to prove the second direction to complete the reduction.            


  Now we know that IDS is NP-hard        


Unfortunately, we don't. We'd need a reduction from SAT to IDS and not the other way around.
To convince yourself of why it makes sense, a reduction from $A$ to $B$ means that if we had an algorithm that solves $B$, we could use it to trivially solve $A$. So, in your case - if we had an algorithm that solves SAT we would be able to solve IDS, but that's not what we want to show. We want to claim that if we could solve IDS then we could solve SAT, which means IDS is "at least as hard as SAT".         

Also, notice that you need a karp reduction and not just a  many-one reduction. That is, a many-one reduction which can be computed in polynomial time (your reduction achieves that)


  Furthermore we have to provide an efficient algorithm which uses non-determinism to show that IDS is a member of NP, correct?


Correct.
Assuming we have provided a karp reduction from SAT to IDS, or equivalently - we proved that $\text{SAT}\le_p \text{IDS}$, then all we know is that IDS is NP-hard, and we need to prove that $\text{IDS} \in \text{NP}$ to get that IDS is NP-complete.