Consider the following coordinate transformation $$y=f(x,t) \qquad t'=t $$ Then by computing partial derivatives using the chain rule: $$\frac{\partial}{\partial x}=\frac{\partial}{\partial y}f_{x} \\ \frac{\partial}{\partial t} = \frac{\partial}{\partial t'}+f_{t}\frac{\partial}{\partial y}$$
I understand that this is correct, but if $t=t'$ then how is it their partial derivatives aren't equal?! What am I missing?
On
The problem is that you cannot consider each coordinate separately to determine how to modify the partial derivatives. This depends on how the chain rule works. To see this consider any smooth function $F(y,t')\rightarrow \mathbb{R}$. Then: \begin{equation} \frac{\partial}{\partial t} {F}(y,t') = \frac{\partial {F}}{\partial y} \frac{\partial y}{\partial t} +\frac{\partial {F}}{\partial t'} \frac{\partial t'}{\partial t} = \left(f_t \frac{\partial}{\partial y} + 1 \frac{\partial}{\partial t'}\right) {F} \end{equation} This shows you that in order to see how ${F}$ changes in $t$ you need to keep track of how it changes with $y$ as $y$ also depends on $t$. For this reason you cannot simply have $\partial /\partial t = \partial /\partial t'$.
When you compute $\partial/\partial t$, it is understood that $x$ should be held constant as $t$ varies, but when you compute $\partial/\partial t'$, it is $y$ that should be held constant as $t'$ varies. And this is not the same thing, since $y \neq x$.