Forcing and new ordinals $\textbf{Question}$: In expositions to forcing, why do we insist on not adding new ordinals to a countable transitive model $M$ of ZFC? 
For example, after ruling out transitive proper class models as possible candidates for satisfying (ZFC plus) $\textbf{V} \neq \textbf{L}$, Kunen writes on pages 185 
"We shall describe a general procedure for finding countable transitive models $N$ of ZFC such that $M \subseteq N$ and $o(M) = o(N)$." 
Why insist on $o(M) = o(N)$? Later, he goes on to give examples of "non-generic" objects that shouldn't be added to $M$ as they code ordinals $> o(M)$ (Exercise (A3)). Why is this an issue?
 A: You might consider the following quote from Cohen's paper, "The Discovery of Forcing", Rocky Mountain Journal of Mathematics, Volume 32, number 4, 2002, pg. 1091 (found under title on the Web):

So we are starting with a countable standard model $M$, and we wish to to adjoin new elements and still obtain a model.  An important decision is that no new ordinals are to be created.  Just as Godel did not remove any ordinals in the constructible universe, a kind of converse decision is made not to add any new ordinals.

Since (as you probably know from your readings about forcing) Prof. Cohen was the creator of the forcing technique, forcing was (at its inception), designed not to add new ordinals.  The natural question to ask now is, "Why?" (this, of course, is the question you already asked).
A clue as to why is found in Jech's Set theory:  Third Millenium Edition, Chapter 13, in Theorem 13.28 and the preceding paragraph:

If $M$ is a transitive model of $ZFC$, then the Axiom of Choice in $M$ enables us to code all sets in $M$ by sets of ordinals and the model is determined by its sets of ordinals.  The precise statement of this fact is:  if $M$ and $N$ are two transitive models of $ZFC$ with the same sets of ordinals, then $M$ = $N$ [in which case, $M$ and $N$ have identical ordinals--my comment]. In fact, a slightly stronger assertion is true (on the other hand, one cannot prove that $M$ = $N$ if neither model satisfies $AC$.)
Theorem 13.28.  Let $M$ and $N$ be transitive models of $ZF$ and assume that the Axiom of Choice holds in $M$.  If $M$ and $N$ have the same sets of ordinals, i.e., $P^{M}$($Ord^{M}$) = $P^{N}$($Ord^{N}$) [$P^{M,N}$ is just the power set operation restricted to the models $M$, $N$, respectively and $Ord^{M,N}$ are just the ordinals of $M$ and $N$, respectively (Kunen uses essentially the same definition for $Ord^{M, N}$)--my comment], $M$ = $N$.

As regards Profs. Hamkins' and Karagila's comments to each other regarding Prof. Hamkins' answer to you, you might ask either (or both) of them for a proof of their assertion, i.e., that adding ordinals to a model of $ZFC$ + $V$ $\ne$ $L$ might (or would) make $V$ = $L$ true again.  If they would design their proofs so as to educate, I'm sure you would find them very interesting (and then you would truly know and understand why one "doesn't insist on adding ordinals to a countable transitive model $M$ of $ZFC$ in expositions of forcing").
As regards Prof. Hamkins' assertion that there are "other model-theoretic methods that do add new new ordinals", you might take a look at his answer to S A's mathoverflow question, "Can 'syntactic forcing' add ordinals?".  As regards the fact that forcing does not add new ordinals, you might take a look at tomasz's mathstackexchange question, "A question about the proof that forcing extensions don't add ordinals", paying particular attention to Halbeisen's text mentioned,  Profs. Caicedo's and Blass's comments, and the theorem in Halbeisen's text that tomasz mentions in his question.  You might find this information at least somewhat helpful in your search for an answer. 
A: We don't insist on this as a prior goal. Rather, it just happens to be the case that forcing does not add new ordinals, whether you would want it to or not. Thus, this is a feature of forcing rather than a goal. Meanwhile, there are other model-construction methods that do add new ordinals.
But meanwhile, as far as getting models of $V\neq L$ is concerned, it suffices to have a nontrivial extension with the same ordinals, for then the larger model will have to satisfy $V\neq L$. 
A: I think Paul Cohen has a decent (retrospective) justification in Section 2, Chapter IV of his book "Set Theory and the Continuum Hypothesis". Check out the Corollary on page 110 and the discussion following it. 
I am not sure why this issue is ignored by other authors.
