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

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

More comments on the quasi-linear case and the general method of characteristics: the quasi-linear Cauchy problem $$ \frac{\partial u}{\partial t}+\sum_{1\le j\le n} a_j(t,x, u)\frac{\partial u}{\partial x_j}=b(t,x,u),\quad u(0,x)=u_0(x). \tag{4}$$ has a linear companion $$ \frac{\partial F}{\partial t}+\sum_{1\le j\le n} a_j(t,x, v)\frac{\partial F}{\partial x_j}+b(t,x,v)\frac{\partial F}{\partial v}=0,\quad F(0,x,v)=v-u_0(x) \tag{5}$$ where $t,x,v$ are independent variables. It is not difficult to solve using the linear method of characteristics outlined above. Then since $\partial F/\partial v=1$ at $t=0$, the equation $ F(t,x,v)=0 $ determines implicitely $v=u(t,x)$ and the expression of derivatives of $u$ in terms of derivatives of $F$, e.g. $ \partial u/\partial x=-\frac{\partial F/\partial x}{\partial F/\partial v} $ imply that $u$ solves the Cauchy problem (4). Here also the notational industry is working full throttle. People would write $$ \dot x=a(t,x,u)\quad \dot u=b(t,x,u)\quad \text{which is } \frac{dx_j}{a_j}=\frac{du}{b},\quad 1\le j\le n. $$