Since I'm not quite sure I understand the question, I'm going to take a few guesses what it might mean. I am not sure that these guesses rise to the level of an answer, but @paulgarrett has suggested that they might, so I'll give it a try.
Note, by the way, that $N$ here is not just any unipotent subgroup, but the unipotent radical of a parabolic subgroup of $G$ that admits $M$ as a Levi component; and $K$ is not just any maximal compact, but what Cartier calls on p. 140 a "special, good, maximal compact subgroup of $G$" (and there is also a condition that is often phrased as something like its being adapted to $M$). Nowadays the terminology is that $K$ is a special parahoric subgroup, or, rather, a parahoric subgroup associated to a special vertex (and note that, if $G$ is not simply connected, it need not actually be maximal!), and the condition of being adapted to $M$ is just the requirement that that special vertex lie in the canonical image of the enlarged building of $M$ in the reduced building of $G$. (The map is not (usually) canonical, but the image is.)
The literal question "why is this true?" seems somewhat like asking why is it true that the Lebesgue measure of $[0, 1]$ is $1$. It's true because we normalise it that way, and there are good reasons to do so, but nothing would break mathematically if we chose a different normalisation and were careful to be consistent everywhere after.
So perhaps you mean, why can we make this normalisation? For each individual normalisation, we can do so because $\Gamma \cap K$ is a compact, open subgroup of $\Gamma$. For any such subgroup, you can always normalise its measure to be $1$. You cannot, of course, always impose different such normalisations simultaneously; that is to say, a statement about the possibility of normalising the measures of several such subgroups is a theorem. (For example, I think it was Waldspurger who observed that you could normalise the Haar measure on a reductive $p$-adic group in such a way that the measures of parahoric subgroups are described in a uniform way in terms of their reductive quotients.)
So perhaps what you mean is, why can we make all of these normalisations simultaneously? The reason is that we are not asking for one measure satisfying several conditions, but for several measures, each satisfying one condition: we ask for a Haar measure on $N$ that assigns measure $1$ to $N \cap K$, a Haar measure on $M$ that assigns measure $1$ to $M \cap K$, and so on.
A Haar measure on a group does not, in general, determine a Haar measure on each closed subgroup, so there is no general sense in which we can ask whether the Haar measure on $N$, for example, "is" the Haar measure on $G$. (This is what I was getting at with my somewhat mystifying comment that "it is meaningless without such a normalisation to ask whether, for example, the Haar measure of $N \cap K$ equals the Haar measure of $M \cap K$.") A Haar measure on $G$ does determine one on each open subgroup, so we can meaningfully (if somewhat sloppily) ask whether the Haar measure on $K$ "is" the one on $G$; and, fortunately, it is, in the sense that they agree on Borel subsets of $K$.
One can, however, hope that we are not choosing our other measures completely randomly. There are several senses in which we are not, and the Iwasawa decomposition gives one. Namely, we have that the multiplication map $M \times N \times K \to G$ is surjective. It is (usually) not a homomorphism, and (usually) not a bijection, but we can nonetheless wonder whether, for $f \in \operatorname C_c^\infty(G)$, there is a relationship between $\int_K \int_N \int_M f(m n k)\mathrm dm\,\mathrm dn\,\mathrm dk$ and $\int_G f(g)\mathrm dg$, where I have written "$\mathrm d\cdot$" for all Haar measures and left it to context to sort out which is which. The normalisation Cartier proposes is, not the only one, but a natural one that makes sure that these two integrals are equal. This is exactly (4) on p. 145. Other pleasant coherencies among these measures are described in the rest of §IV.
