Yes, your understanding it correct. The Laplace-Beltrami depends on the current state $x$ and you have to recompute $L$ if $x$ changes. Therefore you cannot write matix of $\Delta$ without knowing the state $x$.
To expand on solving $\frac{\partial x}{\partial t} = -\Delta x$:
Discretization in space turns $\Delta$ to $L$, discretization in time turns $\frac{\partial x}{\partial t}$ to $\frac{x_{n+1} - x_n}{h}$ and decide if we evaluate the right hand side at time $t_n$ or $t_{n+1}$
Option 1 = Forward Euler: $$ x_{n+1} - x_n = h L_n x_n $$ where $L_n = L(x_n)$, $L$ is considered as a function of the state $x$. As you have mentioned this method can be unstable if we take too big step $h$. Therefore, we often resort to the second option.
Option 2 = Backward Euler: $$ x_{n+1} - x_n = h L_{n+1} x_{n+1} $$ Because $L_{n+1}$ depends on $x_{n+1}$ the above equation is non-linear system, we cannot solve it with a simple linear solve.
The common approach is to Taylor expand $L$ around the point $x_n$, general Taylor expansion around point $x$ is $$ L(x + \Delta x) = L(x) + \nabla L(x) \cdot \Delta x + \dots $$ therefore for $x= x_n$ and $\Delta x = x_{n+1} - x_n$ we get $$ L_{n+1} = L(x_{n+1}) = L(x_n + \Delta x) = L(x_n) + \nabla L(x_n) \cdot \Delta x + \dots $$
The simplest thing to do is to take just the first term of the Taylor expansion, i.e. we replace $L_{n+1}$ with $L_n$. This way, we end up with a linear problem $$ x_{n+1} - x_n = h L_{n} x_{n+1} $$
You can take higher order terms or solve the Backward Euler equation with non-linear solver. However, in your case that is probably unnecessary and the headache of computing the gradient of $L$ is not worth it(I might be wrong here, I do not have hands on experience with solving this equation).