The problem with intuition about cancelling differentials, it isn't safe. And yet, the method of differentials is stupidly successful.
Let me give a standard example of intuitions downfall. First, since partials cancel,
$$ \frac{\partial z}{\partial y}\frac{\partial y}{\partial x}\frac{\partial x}{\partial z} = 1$$
except, it doesn't. Actually, with the right interpretation,
$$ \frac{\partial z}{\partial y}\frac{\partial y}{\partial x}\frac{\partial x}{\partial z} = -1.$$
In particular, we assume $x,y,z$ are related by some level function $F(x,y,z)=0$ then $dF = F_xdx+F_ydy+F_zdz$ thus
$$ \frac{\partial z}{\partial y} = \frac{dz}{dy}\bigg{|}_{dx=0} = -\frac{F_y}{F_z}$$
with more words, if we consider $z$ as a function of $x,y$ then the partial derivative of $z$ whilst holding $x$ fixed is $-F_y/F_z$. Notice, I simply take the total differential of $F$ and solve for $dz/dy$ while setting $dx=0$. This is an example of how the differential notation is naively successful (because, careful application of the implicit function theorem yields the same outcome). Likewise, intuitive calculation with $dx,dy,dz$ yields
$$ \frac{\partial y}{\partial x} = \frac{dy}{dx}\bigg{|}_{dz=0} = -\frac{F_x}{F_y}$$
$$ \frac{\partial x}{\partial z} = \frac{dx}{dz}\bigg{|}_{dy=0} = -\frac{F_z}{F_x}$$
Thus,
$$ \frac{\partial z}{\partial y}\frac{\partial y}{\partial x}\frac{\partial x}{\partial z} = \left(-\frac{F_y}{F_z}\right)\left(-\frac{F_x}{F_y}\right)\left(-\frac{F_z}{F_x}\right) = -1.$$
Getting back to your posed question. Why are there sums of derivatives? Well, in short, because the multivariate function can change in all of its arguments. As the derivative is a linear approximation to the change in the function we have little hope except to see formulas formed from sums of all the possible things which can change the outcome. This is the multivariate chain rule. It accounts for each entry in an entirely symmetrical manner. Ok, these sort of explainations don't settle well with me. The real answer in my estimation is matrix multiplication. The chain-rules really fall out of multiplication of Jacobian matrices which in turn come from the chain-rule in its pure form $D(F \circ G) = DF \circ DG$. But, perhaps this isn't intuition. That said, it is my intuition.
I'll add a little example to explain how the matrix multiplication works together with the Jacobian matrix to capture the chain rule. Suppose $\vec{X}: \mathbb{R}^2_{uv} \rightarrow \mathbb{R}^3_{xyz}$ and $\vec{F} = \langle P, Q, R \rangle : \mathbb{R}^3_{xyz} \rightarrow \mathbb{R}^3$. Here I use the notation $\mathbb{R}^2_{uv}$ to indicate $u,v$ serve as the coordinates. Here you can think of $\vec{X}$ as a parametrization of a surface and $\vec{F}$ as a vector field in three dimensional space. The composition $\vec{F} \circ \vec{X}$ is commonly considered in the calculation of flux of $\vec{F}$ through the surface parametrized by $\vec{X}$. In this case, the Jacobian of $\vec{X}$ is given by
$$ J_{\vec{X}} = \left[ \frac{\partial \vec{X}}{\partial u} |\frac{\partial \vec{X}}{\partial v}\right] = \left[\begin{array}{cc} \partial_u x & \partial_v x \\
\partial_u y & \partial_v y \\
\partial_u z & \partial_v z \end{array} \right]$$
and the Jacobian of $\vec{F}$ is given by
$$ J_{\vec{F}} = \left[ \frac{\partial \vec{F}}{\partial x}|
\frac{\partial \vec{F}}{\partial y}|
\frac{\partial \vec{F}}{\partial z} \right] = \left[
\begin{array}{ccc}
\partial_x P & \partial_y P & \partial_z P \\
\partial_x Q & \partial_y Q & \partial_z Q \\
\partial_x R & \partial_y R & \partial_z R \\
\end{array} \right]$$
Setting $\vec{G} = \vec{F} \circ \vec{X}$ we find from the matrix form of the chain rule that: (suppressing point dependence)
\begin{align} J_{\vec{G}} &= J_{\vec{F}}J_{\vec{X}} \\
&= \left[
\begin{array}{ccc}
\partial_x P & \partial_y P & \partial_z P \\
\partial_x Q & \partial_y Q & \partial_z Q \\
\partial_x R & \partial_y R & \partial_z R \\
\end{array} \right]\left[\begin{array}{cc} \partial_u x & \partial_v x \\
\partial_u y & \partial_v y \\
\partial_u z & \partial_v z \end{array} \right] \\
&=
\left[\begin{array}{c|c}
\partial_x P\partial_u x +\partial_y P \partial_u y + \partial_z P\partial_u z
&\partial_x P\partial_v x +\partial_y P \partial_v y + \partial_z P\partial_v z \\
\partial_x Q\partial_u x +\partial_y Q \partial_u y + \partial_z Q\partial_u z
&\partial_x Q\partial_v x +\partial_y Q \partial_v y + \partial_z Q\partial_v z \\
\partial_x R\partial_u x +\partial_y R \partial_u y + \partial_z R\partial_u z
&\partial_x R\partial_v x +\partial_y R \partial_v y + \partial_z R\partial_v z
\end{array} \right]
\end{align}
For example, in the $(1,1)$ entry we read off:
$$ \frac{\partial G^1}{\partial u} = \frac{\partial}{\partial u} \left[P(x(u,v), y(u,v), z(u,v))\right] =
\frac{\partial P}{\partial x}\frac{\partial x}{\partial u} +
\frac{\partial P}{\partial y}\frac{\partial y}{\partial u} +
\frac{\partial P}{\partial z}\frac{\partial z}{\partial u}
$$
Notice the matrix $J_{\vec{G}}$ contains all $6$ interesting chain rules involving composition of the component functions $P,Q,R$ of $\vec{F}$ composed with the component functions $x,y,z$ of $u,v$.
Let me put it in this way: $u$ is a function of $x,t$, but can be thought of as a function of the new variables $\xi$ and $\eta$ after the change. Vretblad is using the standard physical formalism and keeps the same name for the function $u(x,t)$ and $u(\xi,\eta)$, so we get the (terrible from the mathematical point of view) identity
$$
u(x,t)=u(\xi(x,t),\eta(x,t)).
$$
Derivating both sides wrt $x$ (using the chain rule in the RHS) we get
$$
u_x=\frac{\partial u}{\partial \xi}\underbrace{\frac{\partial\xi}{\partial x}}_{=1}+\frac{\partial u}{\partial \eta}\underbrace{\frac{\partial\eta}{\partial x}}_{=1}=
\frac{\partial u}{\partial \xi}+\frac{\partial u}{\partial \eta}.
$$
Doing it once again and applying the chain rule to both terms in the RHS gives you
$$
u_{xx}=\color{red}{\frac{\partial}{\partial x}\Bigl(\frac{\partial u}{\partial \xi}\Bigr)}+\color{blue}{\frac{\partial}{\partial x}\Bigl(\frac{\partial u}{\partial \eta}\Bigr)}=\color{red}{\frac{\partial^2 u}{\partial \xi^2}\frac{\partial\xi}{\partial x}+\frac{\partial^2 u}{\partial\eta\partial\xi}\frac{\partial\eta}{\partial x}}+\color{blue}{\frac{\partial^2 u}{\partial \xi\partial\eta}\frac{\partial\xi}{\partial x}+\frac{\partial^2 u}{\partial\eta^2}\frac{\partial\eta}{\partial x}}=...
$$
I hope you can continue after that.
Best Answer
In the linked pdf, notice that you have two functions, $F$ and $f$. The function $f$ is an expression in $u$ and $v$ that equals $F$ when you substitute. For example, if $F(x,y) = x - y$, $u = x$, and $v = -y$, you would need $f(u,v) = u + v$.
To maybe see how it can go wrong if you incorrectly assume $F = f$, let $m = n = x$ in your example. You would have: $$ \frac{\partial f}{\partial x} = \frac{\partial x}{\partial x}\frac{\partial f}{\partial x} + \frac{\partial x}{\partial x}\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial x}. $$