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 $$\frac{\partial}{\partial t}+ \sum_{1\le j\le n} a_j(t,x)\frac{\partial}{\partial x_j}. $$