That your G is a test category is stated in the last sentence of 4.1.20 in the paper of Cisinski you mention. This case is also treated in more detail in section 8.3, where it is shown that the left Kan extension/restriction along both adjunctions induced by your "forgetful functor" are Quillen equivalences (8.3.8).
(By the way, if I recall correctly, it is a corollary that if we give simplicial sets the Joyal model structure instead, then both adjunctions are still Quillen pairs, and they realize the two adjoints to the inclusion of (∞,0)-categories in (∞,1)-categories.)
As to "why is the unit interval the canonical interval?", there is an interesting universal property of the unit interval given in some observations of Freyd posted at the categories list, characterizing $[0, 1]$ as a terminal coalgebra of a suitable endofunctor on the category of posets with distinct top and bottom elements.
There are various ways of putting it, but for the purposes of this thread, I'll put it this way. Recall that the category of simplicial sets is the classifying topos for the (geometric) theory of intervals, where an interval is a totally ordered set (toset) with distinct top and bottom. (This really comes down to the observation that any interval in this sense is a filtered colimit of finite intervals -- the finitely presentable intervals -- which make up the category $\Delta^{op}$.) Now there is a join $X \vee Y$ on intervals $X$, $Y$ which identifies the top of $X$ with the bottom of $Y$, where the bottom of $X \vee Y$ is identified with the bottom of $X$ and the top of $X \vee Y$ with the top of $Y$. This gives a monoidal product $\vee$ on the category of intervals, hence we have an endofunctor $F(X) = X \vee X$. A coalgebra for the endofunctor $F$ is, by definition, an interval $X$ equipped with an interval map $X \to F(X)$. There is an evident category of coalgebras.
In particular, the unit interval $[0, 1]$ becomes a coalgebra if we identify $[0, 1] \vee [0, 1]$ with $[0, 2]$ and consider the multiplication-by-2 map $[0, 1] \to [0, 2]$ as giving the coalgebra structure.
Theorem: The interval $[0, 1]$ is terminal in the category of coalgebras.
Let's think about this. Given any coalgebra structure $f: X \to X \vee X$, any value $f(x)$ lands either in the "lower" half (the first $X$ in $X \vee X$), the "upper" half (the second $X$ in $X \vee X$), or at the precise spot between them. Thus, you could think of a coalgebra as an automaton where on input $x_0$ there is output of the form $(x_1, h_1)$, where $h_1$ is either upper or lower or between. By iteration, this generates a behavior stream $(x_n, h_n)$. Interpreting upper as 1 and lower as 0, the $h_n$ form a binary expansion to give a number between 0 and 1, and therefore we have an interval map $X \to [0, 1]$ which sends $x_0$ to that number. Of course, should we ever hit $(x_n, between)$, we have a choice to resolve it as either $(bottom_X, upper)$ or $(top_X, lower)$ and continue the stream, but these streams are identified, and this corresponds to the identification of binary expansions
$$.h_1... h_{n-1} 100000... = .h_1... h_{n-1}011111...$$
as real numbers. In this way, we get a unique well-defined interval map $X \to [0, 1]$, so that $[0, 1]$ is the terminal coalgebra.
(Side remark that the coalgebra structure is an isomorphism, as always with terminal coalgebras, and the isomorphism $[0, 1] \vee [0, 1] \to [0, 1]$ is connected with the interpretation of the Thompson group as a group of PL automorphisms $\phi$ of $[0, 1]$ that are monotonic increasing and with discontinuities at dyadic rationals.)
Best Answer
Yes, you can see this as happening in a "fibrational cosmos." I'll describe how it goes, but then we'll see that the description of $\widehat{C}/H$ that comes out could also be deduced pretty naively.
The universal property of $\widehat{C}$ is that the category of functors $A\to \widehat{C}$ is equivalent to the category of discrete fibrations from $A$ to $C$, via pullback of the universal such. (A discrete fibration from $A$ to $C$ is a span $A\leftarrow E \to C$ such that $E\to C$ is a fibration, $E\to A$ is an opfibration, the two structures are compatible, and $E$ is discrete in the slice 2-category over $A\times C$.)
Now the slice category $\widehat{C}/P$, for $P\in\widehat{C}$, also has a universal property: it is the comma object of $\mathrm{Id}_{\widehat{C}}$ over $P:1\to \widehat{C}$. Thus a functor $A\to \widehat{C}/P$ is equivalent to a functor $A\to \widehat{C}$ and a natural transformation from it to the composite $A \to 1 \overset{P}{\to} \widehat{C}$. By the universal property of $\widehat{C}$, this is the same as giving a discrete fibration $A\leftarrow E \rightarrow C$ together with a map to the discrete fibration classified by $P:1\to \widehat{C}$, which is just $1\leftarrow \int P \rightarrow C$.
Next, (discrete) fibrations have the special property that not only is the composite of two (discrete) fibrations again a (discrete) fibration, but if $g$ is a discrete fibration and $g\circ f$ is a fibration, then $f$ is a fibration. Therefore, given a discrete fibration from $A$ to $C$ together with a map $E\to \int P$ over $C$, the map $E\to \int P$ is itself a fibration, and we can actually show that $A\leftarrow E \to \int P$ is itself a discrete fibration from $A$ to $\int P$, and that this is an equivalence. Hence, the slice category $\widehat{C}/P$ has the same universal property as $\widehat{\int P}$, so they are equivalent.
Now we can ask about replacing $P:1\to \widehat{C}$ with a more general functor $H:D\to \widehat{C}$. Here the fibrational-cosmos argument fails, because in the discrete fibration $D \leftarrow \int H \to C$ it is no longer true that the solitary map $\int H \to C$ is itself a discrete fibration, so cancellability no longer holds and $E\to \int H$ is no longer necessarily a fibration. However, at least in the 2-category $Cat$, we can still use it to figure out what $\widehat{C}/H$ should look like, by looking at the case $A=1$. In this case, to give a map $1 \to \widehat{C}/H$ means to give a map $d:1\to D$ (i.e. an object of $D$) along with a discrete fibration $E\to C$ and a map from $E$ to $D \leftarrow \int H \to C$ over $d$ and $\mathrm{Id}_C$. This is the same as a map from $E$ to $d^*(\int H) = \int H(d)$ over $C$, and we can then apply the previous argument here since both are discrete fibrations over $C$.
Thus, an object of $\widehat{C}/H$ consists of an object $d\in D$ together with a presheaf on $\int H(d)$. This is not a surprise, though, since we're just applying the original fact $\widehat{C}/P \simeq \widehat{\int P}$ objectwise. Since $d$ can vary between objects, what we're saying is really that $\widehat{C}/H$ is the category of elements of the functor $D\to Cat$ defined by $d\mapsto \widehat{\int H(d)}$, or more evocatively $$ \widehat{C}/H = \int_d \widehat{\int H(d)} $$ as a "double integral".