Sorry, but I think much of this is quite misguided. The issue is that you're mixing up functions with different arguments.
As a simpler example, in classical mechanics, the Lagrangian $$L(q, \dot{q}, t)$$ is a function of multiple variables. It doesn't make sense to take a "total derivative" with respect to $t$. However, if we evaluate it on a specific path $\bar{q}(t)$, then we can construct the function of a single variable $$\bar{L}(t) \equiv L(\bar{q}(t), \dot{\bar{q}}(t), t).$$ For example, in the Euler-Lagrange equation $$\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}$$ the partial derivative with respect to $\dot{q}$ involves the function of multiple variables $L$, but the total derivative $d/dt$ involves a function of one variable. In particular, it is completely meaningless to try to speak of the total derivative of $L(q, \dot{q}, t)$ with respect to $t$, if you don't specify a path. Once you do specify a path, it is trivial, because you're left with a function of time alone.
Similarly, when you have $\mathcal{L}(\phi, \partial_\mu \phi, x^\mu)$, it doesn't make sense to take the "total" derivative of $\mathcal{L}$ with respect to $x^\mu$, because it also depends on the field. It only makes sense after you plug in a specific field profile $\phi(x)$ to construct the function $$\tilde{\mathcal{L}}(x) \equiv \mathcal{L}(\phi(x), \partial_\mu \phi(x), t)$$ which then can be differentiated with respect to $x^\mu$. Once you do this, computing $\partial_\mu \tilde{\mathcal{L}}(x)$ is a trivial application of the chain rule. As long as you distinguish $\mathcal{L}$ and $\tilde{\mathcal{L}}$, there's nothing conceptually confusing here.