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 $x^\mu$.

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.

Sorry, but I think much of this is quite misguided. You can only take the total derivative of a function of one variable. Otherwise, if there are multiple variables, you take partial derivatives.

The reason for the confusion is that we often conflate functions of both types. For example, in classical mechanics, the Lagrangian $$L(q, \dot{q}, t)$$ is a function of multiple variables. 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.

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 $x^\mu$.