If $A\neq0$ then the equation (5) is a quadratic equation, and unless your PDE is elliptic ($B^2-4AC<0$), locally it has two solutions $y_1(x)$ and $y_2(x)$ which satisfy Vieta's formula,
$$
A\left(\frac{dy_1}{dx}+\frac{dy_2}{dx}\right)=B,\quad A\frac{dy_1}{dx}\frac{dy_2}{dx}=C.
$$
If your PDE is parabolic ($B^2-4AC=0$) then $y_1=y_2$. It follows that
$$
A\partial_x^2+B\partial_x\partial_y+C\partial_y^2=A\left(\partial_x^2+\left(\frac{dy_1}{dx}+\frac{dy_2}{dx}\right)\partial_x\partial_y+\frac{dy_1}{dx}\frac{dy_2}{dx}\partial_y^2\right)
$$
$$
=A\left(\partial_x+\frac{dy_1}{dx}\partial_y\right)\left(\partial_x+\frac{dy_2}{dx}\partial_y\right)+\mbox{first order terms}.
$$
Therefore (1) now becomes
$$
A\left(\partial_x+\frac{dy_1}{dx}\partial_y\right)\left(\partial_x+\frac{dy_2}{dx}\partial_y\right)u+\mbox{lower order terms}=0,
$$
which is the expected simplification.
This is the high school level answer to the question why characteristics matter. The scientific answer is much longer and involves microlocal analysis (propagation of singularities), quantum field theory (trajectories of classical point particles corresponding to a linear quantum field), etc.
This is, however, not a complete answer to the question posted, for I have no suggestion as to why (4) is relevant.
Edit: By no means could I read the author's mind when they wrote equation (4), but let me elaborate on what could be done along those lines (at OP's request).
(5) is a first order ODE, of which the solutions depend on one parameter (initial data). For every $(x_0,y_0)$ in the domain let $y_1(x;x_0,y_0)$ and $y_2(x;x_0,y_0)$ be the two solutions of (5) satisfying
$$
y_1(x_0;x_0,y_0)=y_0,\quad y_2(x_0;x_0,y_0)=y_0.
$$
In other words, $y_1(x;x_0,y_0)$ and $y_2(x;x_0,y_0)$ are the two characteristics through the point $(x_0,y_0)$. Fix $(x_0,y_0)$ for a moment, and omit their mention for brevity. Let us do with the characteristic $y_1(x)$ first. Consider $P$ and $Q$ as independent functions, then their restriction to the characteristic $P(x,y_1(x))$ and $Q(x,y_1(x))$ as unknown functions of $x$. If $H$ contains no zero order terms then $H(P,Q,x,y_1(x))$ is a function of $P$,$Q$ and $x$. This renders (4) a first order ODE relating two unknown functions $P$ and $Q$ of one variable $x$,
$$
\left[A(x,y_1(x))\frac{dP}{dx}+H(P,Q,x,y_1(x))\right]y_1'(x)+C\frac{dQ}{dx}=0.
$$
If you can solve this equation in one way or another then you obtain
$$
F_1(P,Q,x)=C_1,
$$
where $F_1$ is a function and $C_1$ is a number, both depending on the choice of the characteristics, i.e., $(x_0,y_0)$. Now if you do the same thing with the characteristic $y_2(x)$ you end up with
$$
F_2(P,Q,x)=C_2.
$$
On the other hand, if your $H$ contains zero ordet terms, $H=H(P,Q,x,y,u)$, then
$$
u(x,y_1(x))=\int^x\left[P+y_1'(t)Q\right]dt,
$$
and (4) becomes an integral-differential equation. Good luck solving it.
Now the final solution of (1) depends on the specification of the boundary (Cauchy) data. Suppose that you have a Cauchy curve $\Sigma$ that intersects every characteristic exactly once, and suppose that $P$, $Q$, $R$, $S$, $T$ and $u$ are known along $\Sigma$. Consider only those $(x_0,y_0)\in\Sigma$, and let $p(x_0,y_0)$, $q(x_0,y_0)$ be the values of $P$ and $Q$ at $(x_0,y_0)$. We now recall all dependences upon $(x_0,y_0)$ above. If $(x,y)$ is any point in the domain then let $(x_1,y_1)$ and $(x_2,y_2)$ be the intersections of $y_1(x)$ and $y_2(x)$ with $\Sigma$, respectively,
$$
y=y_1(x;x_1,y_1),\quad y=y_2(x;x_2,y_2).
$$
This brings us to the system
$$
\begin{cases}
y=y_1(x;x_1,y_1)\\
y=y_2(x;x_2,y_2)\\
F_1(P,Q,x;x_1,y_1)=F_1(p(x_1,y_1),q(x_1,y_1),x_1;x_1,y_1)\\
F_2(P,Q,x;x_2,y_2)=F_2(p(x_2,y_2),q(x_2,y_2),x_2;x_2,y_2)
\end{cases}
$$
This is the parametric solution of the problem (1) with given boundary data $p$ and $q$.
If we want an explicit solution, we make a few more steps. The curve $\Sigma$ can be locally parameterized by one variable $s$, so that $(x_1,y_1)=\eta(s_1)$ and $(x_2,y_2)=\eta(s_2)$. Therefore the above is a system of 4 equation for 4 variables $P$, $Q$, $s_1$ and $s_2$. Eliminating $s_1$ and $s_2$ we find the desired solution
$$
P(x,y),\quad Q(x,y).
$$
Finally,
$$
u(x,y)=u(x_0,y_0)+\int_{(x_0,y_0)}^{(x,y)}\left[Pdx+Qdy\right]
$$
along your favourite curve joining $(x_0,y_0)$ and $(x,y)$. The point $(x_0,y_0)\in\Sigma$ therefore we know $u(x_0,y_0)$ from the boundary data.
This is how, I think, you would solve (1) if you want to start from (4). Generally, the method of characteristics always ends with a parametric solution, and for an explicit solution you need to exliminate the dependence upon the characteristic.