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 $u,v$.
If we let $f$ be defined as $f(u,v)=e^{uv}$ instead, for clarity, then
$$
\frac{\partial g}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}
$$
Once you've calculated the first partial derivative, you repeat the above on said partial derivative to get the second derivative.
Best Answer
What is confusing you is that the letters $x$ and $y$ are used for two purposes: (a) As names of the functions $x(t), y(t)$; (b) As names of the "first argument" and "second argument" of $f$. Sadly, this notation is very usual in mathematics, and there is no suitable alternative.
If I could convince to write partial derivatives differently, e.g. instead of $\frac{\partial f}{\partial x}$ you write $f_1'$ ("derivative on the first parameter"), and instead of $\frac{\partial f}{\partial y}$ you write $f_2'$ ("derivative on the second parameter"), then the formula becomes a lot more palatable:
$$z'(t)=\frac{d}{dt}f(x(t),y(t))=f_1'(x(t),y(t))x'(t)+f_2'(x(t),y(t))y'(t)$$
(I also write $x'(t)=\frac{dx}{dt}, y'(t)=\frac{dy}{dt}, z'(t)=\frac{dz}{dt}$, if that helps.)
Now it is obvious we are in no way modifying $x(t)$ or $y(t)$ when calculating $f_1'$ and $f_2'$. Instead, we are keeping constant the second argument of $f$/the first argument of $f$, respectively. After having calculated $f_1'$ and $f_2'$ - both end up as functions of two arguments - you will substitute those arguments with $x(t), y(t)$.
Sadly, this notation I used above is completely non-standard, and for historical reasons we are stuck with the notation that is confusing for you.