This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof.
The proof I'm going to give is due to Bernhard Neumann in An Essay on Free Products of Groups with Amalgamations (Philosophical Transactions of the Royal Society of London, Series A, 246, 919 (1954), pp. 503-554.)
So we have to prove that in a group satisfying
$$ xyx^{-1} = y^2 \qquad (R_1)$$
$$ yzy^{-1} = z^2 \qquad (R_2)$$
$$ zxz^{-1} = x^2 \qquad (R_3)$$
the elements $x$, $y$ and $z$ are trivial.
By inverting $(R_1)$, multiplying on the left by $y$ and on the right by $x$, we get $$yxy^{-1} = y^{-1}x.$$ This easily gives $$y^i x y^{-i} = y^{-i} x \qquad(R_1^{[i]}),$$ for every integer $i$, by induction. The same argument on the second relation gives $$z^i y z^{-i} = z^{-i} y. \qquad (R_2^{[i]})$$
If we now conjugate $(R_3)$ by $y$, the left-hand side becomes
$$\begin{align}yzxz^{-1}y^{-1} &= z^2y\cdot x \cdot y^{-1}z^{-2}\\
& = z^2y^{-1}xz^{-2}\\
& = z^2 y^{-1}z^{-2}\cdot z^2xz^{-2}\\
& = y^{-1}z^2\cdot x^4
\end{align}$$
(the first equality is a double use of the relation $yz = z^2y$, a reformulation of $(R_2)$ ; the second uses $(R_1^{[1]})$ and the last uses the inverse of $(R_2^{[2]})$ and $R_3$ twice).
On the other hand, the left side becomes
$$\begin{align} yx^2y^{-1} &= (y^{-1}x)^2 \\
&= y^{-1}xy^{-1}x \\
&= y^{-3}x^2
\end{align}$$
(the first equality uses the $(R_1^{[1]})$ twice, the last uses the inverse of $(R_1)$).
Put together, we have proven $y^{-1}z^2x^4 = y^{-3}x^2$, which gives
$$z^2 = y^{-2}x^{-2}.\qquad (R^*)$$
If we conjugate $y$ by $z^{-2}$, we now get on the one hand
$$\begin{align}z^{-2}y z^2 &= x^2 y^2 \cdot y \cdot y^{-2} x^{-2} \\
&= x^2 y x^{-2} \\
&= y^4
\end{align}$$
(the first equality uses $(R^*)$ twice, the last uses $(R_1)$ twice.) But, on the other hand, $z^{-2}yz^2 = z^2 y$ because of $(R_2^{[-2]}$). So we finally get $z^2 y = y^4$, which translates to
$$z^2 = y^3.$$
This proves that $y$ and $z^2$ commute. The relation $(R_2)$ then boils down to $z = z^2$, which gives $z = 1$. Because of the symmetries in the presentation, this proves that the group is trivial.
Not very enlightening, but the fact that the corresponding group with 4 generators is highly nontrivial somehow reduces my hopes of ever finding a "good reason" for this group to be trivial.
If the extension is non-split, then relations $r=1$ of $Q$ become $r=w_K(r)$ for some word $w_K(r)$ in the generators of $K$, which have to be determined. You could in principal calculate $w_K(r)$ from the cocycle of the extension.
In your example, we can take $C_2 = \langle z \mid z^2 = 1 \rangle$. In fact $w_K(r)=z$ for each relation $r=1$ of $A_5$ works (but there are other possible choices). Then you get
$$\langle x,y,z \mid x^5=y^2=(xy)^3=z, z^2=1, xz=zx, yz=zy \rangle$$ as a presentation of the binary icosahedral group.
In fact the relations $xz=zx$ and $yz=zy$ are redundant, because $z$ is a power of both $x$ and of $y$, so you can leave those out if you want to.
Best Answer
Notice that $F(S)$ is a group, $N(R) \trianglelefteq F(S)$, $N(R \cup R') \trianglelefteq F(S)$, and $N(R) \subseteq N(R \cup R')$. By the third isomorphism theorem, $$ \frac{F(S)}{N(R \cup R')} \cong \frac{F(S)/N(R)}{N(R \cup R')/N(R)} $$ or, as presentations, $$ \langle S \mid R \cup R' \rangle \cong \frac{\langle S \mid R \rangle}{N(R \cup R')/N(R)} \text{.} $$ Are you able to show that the only part of $\langle S \mid R \rangle$ that is actually sent to the identity by the quotient on the right is $N(R')$? (In particular, the implicit quotient in $\langle S \mid R \rangle$ has already sent all of $N(R)$ to the identity, so only a subset of $N(R')$ remains to be sent there.)