The assumption “ZF is consistent” I studied the Ebbinghaus's logic textbook, which is done under ZF (sometimes is ZFC) set theory, so the Gödel Completeness Theorem is valid in ZF(C) (in my opinion).
When I studied Jech's set theory, it is  going to show that

If ZF is consistent, then so is ZFC+GCH.

Nonetheless, can we really assume that ZF is consistent? By the Gödel Completeness Theorem, if ZF is consistent, then it is satisfiable and so there is a set V collecting all sets in the universe of ZF, which seems a contradiction.
Didn't we immediately get a contradiction when we say “ZF is consistent”? 
How should I understand the assumption “ZF is consistent”?
 A: Well. How can you assume that $\sf PA$ is consistent, if by Gödel's work you can prove its not provable? The answer is that adding assumptions makes your theory stronger. And this is also a way to measure the strength of a theory.
So $\sf ZF$ is stronger than $\sf PA$ because it proves that $\sf PA$ is consistent. And $\sf ZF+\operatorname{Con}(ZF)$ is even stronger, as that theory proves that $\sf ZF$ is consistent. But it is not strong enough to prove $\operatorname{Con}(\sf ZF+\operatorname{Con}(ZF))$. For that we need an even stronger theory.
Judging by the way your question is phrased, your view on this is somewhat Platonist. The universe. If you are taking a Platonistic view on this, then by Gödel's theorem, if you assume that $\sf ZF$ is "a base theory for the universe", then you're essentially saying "I don't know all the true statements in the universe", because you are limited by this theorem. In particular, you don't know whether or not $\operatorname{Con}\sf (ZF)$ is true or false. Thus, you need to put this as an explicit assumption. And while it is true that this assumption itself is not provable from $\sf ZF$, unless $\sf ZF$ is inconsistent, all you do is strengthen your theory.
As Alex Kruchman wrote, a model of $\sf ZF$ is just a set, maybe even a countable set, with a binary relation. It just happens to satisfy the axioms of $\sf ZF$. This set has no "access" to the rest of the universe, and it does not know that it itself is a set, or that there are other sets out there. So yes, while this is a stronger assumption than $\sf ZF$ itself can prove, it's not a terrible assumption.
Even more so, in studying foundational theories, we care about the relation between these theories. Namely, if one of them is consistent, can we prove another is consistent? Or maybe we cannot? That tells you what are the limitations of these theories. The theorem by Gödel that you cite, about $\sf GCH$, simply shows that $\sf ZF$ cannot disprove the axiom of choice, or $\sf GCH$. 
A: You wrote:

Nonetheless, can we really assume that ZF is consistnet? By the Gödel Completeness Theorem, if ZF is consistent, then it is satisfiable and so there is a set V collecting all sets in the universe of ZF, which seems a contradiction.

I added the bold to highlight the key misunderstanding, which I think has been overlooked in the comments. 
If we assume ZF is consistent, then Gödel's Completeness Theorem gives us a model $\mathcal{M}$ of ZF. What is a model?  $\mathcal{M}$ is just a set $M$ given together with a binary relation $E\subseteq M^2$ such that $(M,E)\models \text{ZF}$. It is not the case that $M$ contains "all sets in the universe of ZF" - as you know, there is no universal set. And we don't even know that $E$ has anything to do with the real $\in$ relation on the elements $M$. $M$ is just a bag of stuff, and some of that stuff is related to other of that stuff by the $E$ relation, in such a way that all the axioms of ZF are satisfied. In fact, by the Löwenheim-Skolem theorem, if ZF is consistent, then it has a countable model. 
So I've addressed the "no universal set" concern from the outside: the model $\mathcal{M}$ produced by the completeness theorem doesn't contain all the sets in our ZF universe. But you might also have been concerned about "no universal set" from the inside. That is, how can the set $M$ contain within it a whole universe of a model of ZF, when ZF proves that there is no universal set?
The answer is that while on the outside, we can see that $M$ is a set (maybe even a countable set), from the point of view of $\mathcal{M}$, our mini ZF universe, $M$ is not a set. That is, $\mathcal{M}$, being a model of ZF, satisfies the sentence $\lnot \exists x\forall y (y\in x)$ ("there is no set which contains every set"). This just means that there is no element $a$ of $M$ such that for every element $b$ of $M$, $(b,a)\in E$. The sethood of $M$ in our big ZF universe has nothing to do with it.
