Consider $u_t+cu_x=0$.
Here on p. 4, it is said that the corresponding Lax pair is given by $$ L=\frac{\partial^2}{\partial x^2}-u I,\quad M=-c\frac{\partial}{\partial x} $$ and that $$\frac{\partial L}{\partial t}=-\frac{\partial u}{\partial t}$$
Sorry, but why is $\frac{\partial L}{\partial t}=-\frac{\partial u}{\partial t}$?
If I apply $L$ on some $v$ and differentiate w.r.t $t$ afterwards, I get $$ \frac{\partial}{\partial t}L(v)=\frac{\partial}{\partial t}(v_{xx}-uv)=v_{txx}-u_tv-uv_t $$
so what about the term $v_{txx}-uv_t$?
In the context of Lax pairs, you consider $\partial_tL$ as the derivative of the operator, not of its evaluation on any function $v$. For instance, $\partial_t (\partial_x)$ is taken to be the zero operator, as $\partial_x$ is an operator that does not explicitly depend on the variable $t$.
Indeed, let us look already at eq. (2.1a) and the first line in (3) in your link. When the derivative of $L\psi$ in $t$ is performed, it is computed as $$ \partial_t(L\psi)=(\partial_tL)\psi+L(\partial_t\psi). $$
So the additional contributions coming from the fact that $\psi$ itself depend on $t$ are considered separately. It can maybe help to note that in this equation, $\partial_t$ means something different in the various terms: for the left side and the last term in the right side it is a genuine derivative in $t$ of a function, whilst in the term in the middle it is the derivative of the operator.
To answer your question proper then, the term $v_{txx}-uv_t$ is precisely $L(\partial_tv)$.
If you prefer, you may even take this as a definition for the derivative of the operator: $$ (\partial_tL)\psi:=\partial_t(L\psi)-L(\partial_t\psi) $$ (as one does for instance with the Lie derivative of tensors, if this geometric analogy means anything to you).