Just read the tableau from bottom to top,
\begin{align}
y_{+1}&=y+h\Phi(x,y,h)\\
&=y+\frac h4 f(x,y)+\frac{3h}4 f\left(x+\frac{2h}3,y+\frac{2h}3 f\left(x+\frac{h}3,y+\frac{h}3f\left(x,y\right)\right)\right)
\end{align}
In the methods that are in the intersection of Heun's and Kutta's approaches, that is, $a_{i,i-1}=c_i$, $a_{i,j}=0$ for $j\ne i-1$, you get a simple nested expression like this.
There just are not many methods of that class, classical RK4 and some lower order methods.
In the 3-stage case
\begin{array}{l|lll}
0\\
c_2&c_2\\
c_3&0&c_3\\
\hline
&b_1&b_2&b_3
\end{array}
the order conditions reduce to
$$
I)~b_1+b_2+b_3=1,~~
II)~b_2c_2+b_3c_3=\frac12,~~
III)~b_2c_2^2+b_3c_3^2=\frac13,~~
\text{and}~~
IV)~b_3c_3c_2=\frac16.
$$
so that (II&IV) $b_2c_2^2=\frac12c_2-\frac16$, (..&III) $b_3c_3^2+\frac12c_2=\frac12$ and then (..&IV) $\frac16c_3=\frac12c_2(1-c_2)$, (&IV) $b_3c_2^2(1-c_2)=\frac1{18}$. Thus one has one degree of freedom to choose $c_2\in (0,1)$, the other coefficients then follow,
$$
c_3=3c_2(1-c_2),~b_3=\frac1{6c_2c_3},~b_2=\frac1{2c_2}-\frac1{6c_2^2},~b_1=1-b_2-b_3.
$$