So from the second last line to the last line, I don't understand how we can deduce that equivalence. In my notes, it says it comes straight from the definition of total differentials, but I don't see how. Could someone perhaps explain the implication with a few more lines in between? Thanks.
It seems that the misunderstanding comes from the confusion between $u(x,y)$ and $U(x,t)=u(x,y(t))$
Generally, the same symbol is used for both because $u(x,y)=U(x,t)$, but it's better to distinguish them until to be familiar to partial derivatives.
For example :
$u(x,y)=xy^2$ and $y(t)=\sin(t)\quad\to\quad U(x,t)=x\sin^2(t)$
The function $U$ is defined any symbol used, for example $U(z,y)=z\sin^2(y)$ which is different from $u(z,y)=zy^2$.
So, for partial differentiation, it is important to make clear which function $u(x,y)$ or $U(x,t)$ is differentiated with respect to what variable.
In your second last line to the last line, $\frac{\partial u}{\partial y}$ must be understood as a partial derivative of $u(x,y)$ which is obvious.
In your last line, $\frac{\partial u}{\partial x}\bigg|_t$ must be understood as a partial derivative of $U(x,t)$, that is $$\frac{\partial u}{\partial x}\bigg|_t=\frac{\partial U}{\partial x}$$ and $\frac{\partial u}{\partial t}\bigg|_x$ must be understood as a partial derivative of $U(x,t)$, that is $$\frac{\partial u}{\partial t}\bigg|_x=\frac{\partial U}{\partial t}$$ This isn't a direct partial derivative of $u(x,y)$ because there is no $t$ in this function, while $t$ is in the function $U(x,t)$.
