Modifying the problem. Rather than consider the PDE $u_{xt}+u\ u_{xx}+\frac{1}{2}u_{x}^2=0$ with initial condition $u(x,0)=u_0(x)$ as asked above, I will consider the following variant.
$$
\text{Solve }u_{xy}+u\ u_{xx} + u_x^2=0\text{ subject to }u(x,0)=f(x).\qquad(\star)
$$
There are three differences between this question and that which was asked originally.
- The coefficient of $u_x^2$ has changed from $\frac{1}{2}$ to $1$.
- The variable $t$ has been renamed to $y$.
- The initial function $u_0(x)$ has been renamed to $f(x)$.
Only (1) represents a significant modification of the problem. It makes the solution more tractable and enables it to be found using an elementary application of the method of characteristics. For these reasons, it is conceivable that this was the intended question.
Note: I will not delve into regularity of the solutions in this answer.
Reduction to a first order quasilinear PDE. Write the equation as
$$
\frac{\partial}{\partial x}\left(u_y+u\ u_x\right)=0.
$$
Thus $(\star)$ is equivalent to
$$
u_y+u\ u_x=g(y),\qquad u(x,0)=f(x),\qquad (\star\star)
$$
where $g(y)$ is an arbitrary function of $y$ (with sufficient regularity).
Method of characteristics.
Perhaps the simplest formulation of the method of characteristics is for quasilinear first order PDEs. These are PDEs of the form
$$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u).$$
To solve this equation, one regards the solution as a surface $z=u(x,y)$ in $xyz$-space. Let $s$ parametrize the initial curve $\bigl(s,0,f(s)\bigr)$ and let $t$ be a second parameter, which can be thought of as the distance flowed along a characteristic curve emanating from $\bigl(s,0,f(s)\bigr)$.
The characteristic equations are then
$$
\frac{dx}{dt}=a(x,y,z),\quad \frac{dy}{dt}=b(x,y,z),\quad \frac{dz}{dt}=c(x,y,z).
$$
Returning to our equation $(\star\star)$, this reduces to $a(x,y,u)=u$ and $b(x,y,u)=1$ and $c(x,y,u)=g(y)$.
Thus
$$
\frac{dx}{dt}=z,\quad \frac{dy}{dt}=1,\quad \frac{dz}{dt}=g(y)
$$
with initial conditions $x(0)=s$ and $y(0)=0$ and $z(0)=f(s)$.
The solution to this system is
$$
x=s+zt,\quad y=t,\quad z=f(s)+h(t),
$$
where $h(t)$ is the antiderivative of $g(t)$ satisfying $h(0)=0$. Since $g$ was arbitrary, so is $h$ given $h(0)=0$.
The solution. Now we eliminate all occurrences of $t$ by replacing them with $y$, then eliminate $s$ by writing $s=x-zy$. Finally, replace $z$ with $u$ to obtain the implicit equation
$$
\boxed{u=f(x-uy)+h(y)},
$$
where $h(y)$ is any sufficiently regular function satisfying $h(0)=0$. This is an implicit equation for the general solution of $(\star)$.
TL;DR. Change the $\frac{1}{2}$ in the original question to $1$ to obtain a PDE solvable by the method of characteristics.
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.
Best Answer
First family: “This allows me to divide both sides by $x^2$.” No, you can't. The variables $x$ and $t$ are not independent, so you can't take $x^2$ outside of the $t$ integral.
What you have is $$dt/t^2= dx/x^2 \iff d(-1/t)=d(-1/x) \iff d(1/t-1/x)=0 \iff 1/t-1/x=c_1.$$
Second family: If $A=B$, then $(tA+xB)/(t+x)=(tA+xA)/(t+x)=A(t+x)/(t+x)=A$ (which also equals $B$). That is, $$ A=B \quad\implies\quad A = B = \frac{tA+xB}{t+x} . $$ The computation starts out by applying this identity with $A=dt/t^2$ and $B=dx/x^2$.
And then you get $$ dt/t + dx/x = du/u \iff d \ln|t| + d \ln|x| = d \ln|u| \iff d \ln|\tfrac{u}{xt}|=0 \iff \tfrac{u}{xt}=c_2 . $$ (If the logarithm of the absolute value of something is constant, then that something itself is constant too.)