Let me work with $n$ dimensions: you want to study the vector field
$$
X=\sum_{1\le j\le n} a_j(x)\frac{\partial}{\partial x_j},
\tag {1}$$
and in particular find the so-called first integrals of $X$ i.e. the functions $f$ such that $Xf=0$. You introduce the system of ODE:
$$
\dot x(t,y)=a(x(t,y)),\quad x(0,y)=y.
\tag {2}$$
The solutions $t\mapsto x(t,y)$ are the integral curves of $X$.
You realize easily that a function is a first integral iff it is constant along the integral curves of $X$: just compute
$$
\frac{d}{dt}\bigl(f(x(t,y))\bigr)=\sum_{1\le j\le n} \frac{\partial f}{\partial x_j}(x(t,y))a_j(x(t,y))=(Xf)(x(t,y))
$$
It means that solving the PDE (1) is somehow equivalent to solving (2). 

Now the notational business. It is tempting to write (2), which is  $
\frac{dx_j}{dt}=a_j(x), 1\le j\le n,
$
symbolically as
$$
\frac{dx_1}
{a_1(x)}=\dots=\frac{dx_n}
{a_n(x)}
$$
since they are all equal to $dt$ ! Well just take this as a symbolic notation which eliminates the presence of the parameter $t$.

Now the Cauchy problem for this autonomous vector field $X$: find an hypersurface $\Sigma$ to which $X$ is transverse, i.e. $X$ is not tangent to $\Sigma$. Then the Cauchy problem
$$
\begin{cases}
Xu=f,\quad \\
u_{\vert \Sigma}=g
\end{cases}
$$
has locally a unique solution: this problem is equivalent to the scalar ODE
$$
\frac{d}{dt}\bigl( u(x(t,y))\bigr)=f(x(t,y)),\quad u(x(0,y))=u(y)=g(y) \text{ for $y\in \Sigma$},
$$
so that
$$
u(x(t,y))= u(y)+\int_0^tf(x(s,y)) ds\quad \text{ for $y\in \Sigma$}.
\tag{3}$$
Note that $y$ moves on $\Sigma$ ($(n-1)$ degree of freedom) and $t$ in $\mathbb R$ so that it is a nice choice of coordinates to pick $y\in \Sigma$ and $t\in \mathbb R$.

There are variants of this when the vector field is not autonomous, i.e. is of type
$$\frac{\partial}{\partial t}+
\sum_{1\le j\le n} a_j(t,x)\frac{\partial}{\partial x_j}.
$$