$$f(x(z,w),y(z,w),z,w)=g(z,w)$$
Don't confuse $\frac{\partial f}{\partial z}$ with $\frac{\partial g}{\partial z}$
$\frac{\partial f}{\partial z}$ means the partial derivative of $f$ with respect to $z$ for $x$ independant of $z$, which is not the same for $g$.
Thus the correct expression is :
$$\frac{\partial g}{\partial z} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial z}+ \frac{\partial f}{\partial z}$$
It is clear that $$\frac{\partial g}{\partial z} \neq \frac{\partial f}{\partial z}$$
However, loosely the same symbol $f$ is used instead of two distinct symbols $f$ and $g$. This is acceptable for people familiar with this ambiguous writting. They are aware of apparent contradictions which can arise and they avoid mistakes in mentally making the distinction between the two different signification of the symbol $f$.
IN ADDITION :
Since this seems difficult to well understand, consider a concret example :
$$f(x,y,z,w)=5x+4y+3z+2w$$
$$\frac{\partial f}{\partial x}=5\quad;\quad \frac{\partial f}{\partial y}=4\quad;\quad \frac{\partial f}{\partial z}=3$$
Then consider another function $g(z,w)$ defined from the preceeding function in the particular case of
$$x(z,w)=6z+7w\quad\text{and}\quad y(z,w)=8z+9w$$
$$\frac{\partial x}{\partial z}=6\quad;\quad \frac{\partial y}{\partial z}=8$$
$$g(z,w)=5(6z+7w)+4(8z+9w)+3z+2w$$
$$g(z,w)=65z+73w$$
Obviously $5x+4y+3z+2w$ is something else than $65z+73w$ . So $f(x,y,z,w)$ and $g(z,w)$ are not a same function. They cannot be confused.
$$\frac{\partial g}{\partial z}=65$$
We see that $\frac{\partial g}{\partial z}$ is obtained directly by partial differentiation of $g(z,w)$. But instead of, if we want to compute $\frac{\partial g}{\partial z}$ indirectly from the properties of the functions $f(x,y,z,w)$ , $x(z,w)$ and $y(z,w)$ we apply the chain rule of derivations :
$$\frac{\partial g}{\partial z} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial z}+ \frac{\partial f}{\partial z}$$
$$\frac{\partial g}{\partial z} =5*6+4*8+3=65$$
As expected, the same result as before $65$ is obtained without expressing explicitely $g(z,w)$. This is purely a consequence of the chain rule. There is no need for more explanation insofar the theory of the chain rule was studied and understood.
Best Answer
The issue is that you are using $x$ as both as a free variable, and as a function, $x(t)$. And likewise $y$.
The notation $\dfrac{\partial f}{\partial t}$ is really being used as an implicit shorthand for $\dfrac{\partial f(x(t),y(t))}{\partial t}$.
$~$ Where $f(x(t),y(t))$ is a convolution of the bivariate function $f$, with monovariate functions $x$ and $y$, each evaluated with the same argument $t$, a free variable.
Likewise you are using the notation $\dfrac{\partial f}{\partial x}$ is used as shorthand for $\left.\dfrac{\partial f(u,v)}{\partial u}\right\vert_{\raise{2ex}{u:=x(t)\\v:=y(t)}}$ .
$~$ That means to evaluate the partial differential of $f$ with respect to its first argument and then form a composition by substituting those arguments with $x(t)$ and $y(t)$.
$~$ Thus you are evaluating the partial differential of the field $f(u,v)$ over the $t$ parametised curve $\{\langle x(t),y(t)\rangle\}$
So the chain rule is actually:
$$\dfrac{\partial f(x(t),y(t))}{\partial t}=\left.\dfrac{\partial f(u,v)}{\partial u}\right\vert_{\raise{2ex}{u:=x(t)\\v:=y(t)}}\cdotp\dfrac{\partial x(t)}{\partial t}+\left.\dfrac{\partial f(u,v)}{\partial v}\right\vert_{\raise{2ex}{u:=x(t)\\v:=y(t)}}\cdotp\dfrac{\partial y(t)}{\partial t}$$
But this is annoying to read and write, so the shorthand is used for convenience.