Factoring Anderson's RSA backdoor
does starting the iteration from $k=1$ instead of $k=P'$ make a difference?


If you start iteration from $k=P'$ then you get;
$$P = A\cdot (P'+i) + P'$$
where $ i = 0,1,2,\ldots$. Take modulo $A$
$$P  = A\cdot (P'+i) + P' \pmod A$$
$$P  =  P' \pmod A$$
Therefore it will still work to reveal the $P'$


in the way $N'$ is generated, what is the best way to factor $N'$ given the information how it is generated? Is there a certain factoring algorithm that makes factoring $N'$ very easy?


The current factoring record in the open literature is 829-bit, however, the recommended key size is at least 2048, i.e each factor has 1024-bits. Therefore the size of $A$ must be around $1024$-bit. Once you set $A$, then you can seek $P = A\,k_1 + P'$ where $P$ is prime.
There is nothing that prevents one to generate small primes for $P'$  and $Q'$. The $P'$ and $Q'$ doesn't add much to $P = A\,k_1 + P'$ since they are added. The $A$ and $k_1$ is important.
Therefore you can choose $P'$ and $Q'$ as below as $829$-bits.
You can use the CADO-NFS to factor the $P'\cdot Q'$. Here some notes about it, too.