Why can't a class model serve as a witness of the consistency of a theory? In the book London Mathematical Society Student Texts 98, or Fast Track to Forcing, there is such a comment in the footnote on the fact that the Von Neumann universe satisfies all axioms of ZFC:

This is not to say that we found a model of ZFC, as V is not a set, it is a proper class.

I can't figure out why a class model can't serve as a witness of the consistency. In fact, I think a class model is fine:
'A theory who has a model is consistent' is a result of the soundness of the deduction system. Checking the proof of the soundness, say the one in Enderton's A Mathematical Introduction to Logic, I failed to find anything like a requirement for the universe of a model to be a set, i.e., the soundness does not require the universe of a model to be a set. That is to say, even if we define a model to include those proper class models, the statement 'a theory who has a model is consistent' is still true.
So, what's wrong with a class model?
 A: This is a very subtle point which requires careful understanding of precisely how the basic definitions and proofs about the satisfaction relation in a model work (and in particular, how recursion is implemented rigorously in set theory).  Given a model $M$, you define a relation $M\models\varphi(a)$ on pairs $(\varphi,a)$ where $\varphi$ is a formula in the language of $M$ and $a$ is a tuple of elements of $M$ to be substituted for the free variables of $\varphi$.  This definition is done by recursion on the complexity of $\varphi$.
But, how do you know that definitions by recursion actually work?  What you do is prove by induction that each initial segment of the definition by recursion exists.  That is, you prove by induction on $n$ that for each $n\in\mathbb{N}$, there exists a unique relation $M\models_n\varphi(a)$ where $\varphi$ is restricted to the set of formulas of length at most $n$ which satisfies the recursive definition of satisfaction.  The general satisfaction relation is then the union $\bigcup_{n\in\mathbb{N}}\models_n$.
Now, what goes wrong if $M$ is a proper class?  Well, note that these relations $\models_n$ are themselves proper classes, since they are relations on pairs $(\varphi,a)$ where $a$ is a tuple of elements of $M$ (so there is a proper class of values that $a$ can take).  This means that in ZFC, you cannot express the statement "for each $n\in\mathbb{N}$, there exists a unique relation $M\models_n\varphi(a)$...", since it involves quantifying over classes.  So this recursive definition cannot be carried out, and you cannot even define the relation $\models$ on a proper class, let alone prove (or even express) that proofs are sound with respect to it.

Now given all of this, you probably are wondering: if satisfaction cannot be defined for proper class structures, what does it even mean to have a "class model"?  The answer is that although you cannot carry out the recursive definition to define the relation $\models$ on the class of all pairs $(\varphi,a)$, for any individual formula $\varphi$ you can still define $M\models\varphi(a)$ by just carrying out the finitely many steps of the recursion "by hand".  For instance, if $\varphi$ is $x\in y$, you can define $M\models \varphi(a,b)$ to mean "$a\in_M b$", and you can define $M\models\forall x(\varphi(a,x))$ to mean "$\forall x\in M(a\in_M x)$".
So, for each individual axiom $\varphi$ of ZFC, you can say what it means for the class $V$ to be a model of $\varphi$.  When we say "$V$ is a class model of ZFC", this is really a meta-theory statement: it is saying we have a meta-theorem that says for each axiom $\varphi$ of ZFC, "$V\models \varphi$" is a theorem.  Crucially, this meta-theorem cannot be expressed as a single ordinary theorem, since "$V\models\varphi$" is not a formula in which $\varphi$ appears as a free variable (so that we can quantify over it and say it is true for all $\varphi$).  Instead, it is a separate sentence for each specific choice of $\varphi$, which cannot be combined into a single quantified sentence.
In the same way, for each individual proof, you can show that this notion of satisfaction is sound with respect to that proof.  So, if you had some specific proof of a contradiction from ZFC, you could implement that proof in your class model $V$ to obtain a contradiction.  But this shouldn't be a surprise: if you have a specific proof of a contradiction from ZFC, then of course you can reach a contradiction, since you can just carry out that proof of a contradiction!  What you can't do is prove a general theorem that says if there is any proof of a contradiction from ZFC, then $0=1$.  This is again because the argument above is a separate argument for each individual proof, and so it cannot be turned into a single universally quantified theorem about all proofs.
A: The big issue with proper-class-sized structures is that they're too large in general to "use $\models$ properly." This is a bit subtle, so let's look at an example:
Let's use a "sufficiently strong" finitely axiomatizable subtheory $T$ of $\mathsf{ZFC}$ instead of $\mathsf{ZFC}$ itself (basically this means that we don't have to worry about "nonstandard axioms" - if this isn't something you're familiar with, ignore it for now, it's inessential). Godel's second incompleteness theorem applies to $T$, so there is a model $\mathcal{M}$ of $T+\mathsf{\neg Con}(T)$. At the same time, $T$ proves the soundness theorem, so $\mathcal{M}$ thinks "if $T$ has a set-sized model then $T$ is consistent." Consequently $\mathcal{M}$ thinks that $T$ has no set-sized model. On the other hand, it's clear from this example that $T$ can't prove "if $T$ has a class-sized model then $T$ is consistent."
At first glance this might suggest that there are two forms of soundness - "set-sized soundness" and "class-sized soundness" - and that the latter is stronger than the former. However, the reality is different: "class-sized soundness" cannot even be expressed in the language of set theory! This is because $\models$ involves quantifying over (a small portion of) the powerset of the structure involved; once the structure we're looking at is big enough, this can't be done directly at all. (An interesting variant of this situation occurs in the context of $\mathsf{NF}$, an alternate set theory where size is not directly relevant to sethood; see here.)
If you want to rigorously talk about class-sized structures, you need to be working in a class theory. At that point it's hyperclass (or conglomerate) sized structures which don't behave properly, and so forth.
