OP is essentially asking the following.

Why the total time derivative $$ \frac{d}{dt}~=~\frac{\partial}{\partial t} + \sum_{m=0}^{\infty}\sum_{i=1}^n q^{i(m+1)}(t)\frac{\partial}{\partial q^{i(m)}(t)} \tag{1} $$ and the functional/variational derivative $$ \frac{\delta}{\delta q^j(t^{\prime})} \tag{2}$$ commute?

That's a good question. The intuitive reason is that the differentiations refer to different variables. But actual calculations make it less obvious (cf. e.g. eq. (3) below). Assume that the derivatives (1) & (2) act on some space ${\cal F}$.

1. For instance, say that ${\cal F}$ is the space of functions of the form $f(q^{i(m)}(t),t)$. Then we may write the functional derivative (2) as $$ \frac{\delta}{\delta q^j(t^{\prime})} ~=~ \sum_{\ell=0}^{\infty}\delta^{(\ell)}(t\!-\!t^{\prime})\frac{\partial}{\partial q^{j(\ell)}(t)},\tag{2'}$$ because an infinitesimal variation is of the form $$\begin{align}\int\! dt^{\prime} \sum_{j=1}^n \frac{\delta f(q^{i(m)}(t),t)}{\delta q^j(t^{\prime})}\delta q^j(t^{\prime}) &~=~\delta f(q^{i(m)}(t),t) ~=~\sum_{j=1}^n \sum_{\ell=0}^{\infty}\frac{\partial f(q^{i(m)}(t),t)}{\partial q^{j(\ell)}(t)}\delta q^{j(\ell)}(t)\cr &~=~\int\! dt^{\prime} \sum_{j=1}^n \sum_{\ell=0}^{\infty}\delta^{(\ell)}(t\!-\!t^{\prime})\frac{\partial f(q^{i(m)}(t),t)}{\partial q^{j(\ell)}(t)} \delta q^j(t^{\prime}). \end{align} \tag{2"} $$ When we calculate the commutator $$\begin{align}\left[\frac{\delta}{\delta q^j(t^{\prime})}, \frac{d}{dt}\right] &\stackrel{(1)+(2')}{=}~ \sum_{m=0}^{\infty}\sum_{i=1}^n \left[\frac{\delta}{\delta q^j(t^{\prime})},q^{i(m+1)}(t)\right]\frac{\partial}{\partial q^{i(m)}(t)} -\sum_{\ell=0}^{\infty}\left[\frac{d}{dt}, \delta^{(\ell)}(t\!-\!t^{\prime})\right]\frac{\partial}{\partial q^{j(\ell)}(t)} \cr &~~~=~~~ \sum_{m=0}^{\infty} \delta^{(m+1)}(t\!-\!t^{\prime}) \frac{\partial}{\partial q^{i(m)}(t)} - \sum_{\ell=0}^{\infty} \delta^{(\ell+1)}(t\!-\!t^{\prime})\frac{\partial}{\partial q^{j(\ell)}(t)}~=~0,\end{align} \tag{3} $$ we get zero!

2. If we extend ${\cal F}$ with functions of finite many other times, or with functionals (say, with internal time-integrations), or both, similar calculations show that the derivatives (1) & (2) commute.

OP is essentially asking the following.

Why the total time derivative $$ \frac{d}{dt}~=~\frac{\partial}{\partial t} + \sum_{m=0}^{\infty}\sum_{i=1}^n q^{i(m+1)}(t)\frac{\partial}{\partial q^{i(m)}(t)} \tag{1} $$ and the functional/variational derivative $$ \frac{\delta}{\delta q^j(t^{\prime})} \tag{2}$$ commute?

That's a good question. The intuitive reason is that the differentiations refer to different variables. But actual calculations make it less obvious (cf. e.g. eq. (3) below). Assume that the derivatives (1) & (2) act on some space ${\cal F}$.

1. For instance, say that ${\cal F}$ is the space of functions of the form $f(q^{i(m)}(t),t)$. Then we may write the functional derivative (2) as $$ \frac{\delta}{\delta q^j(t^{\prime})} ~=~ \sum_{\ell=0}^{\infty}\delta^{(\ell)}(t\!-\!t^{\prime})\frac{\partial}{\partial q^{j(\ell)}(t)},\tag{2'}$$ because an infinitesimal variation is of the form $$\begin{align}\int\! dt^{\prime} &\sum_{j=1}^n \frac{\delta f(q^{i(m)}(t),t)}{\delta q^j(t^{\prime})}\delta q^j(t^{\prime})\cr ~=~&\delta f(q^{i(m)}(t),t)\cr ~=~&\sum_{j=1}^n \sum_{\ell=0}^{\infty}\frac{\partial f(q^{i(m)}(t),t)}{\partial q^{j(\ell)}(t)}\delta q^{j(\ell)}(t)\cr ~=~&\int\! dt^{\prime} \sum_{j=1}^n \sum_{\ell=0}^{\infty}\delta^{(\ell)}(t\!-\!t^{\prime})\frac{\partial f(q^{i(m)}(t),t)}{\partial q^{j(\ell)}(t)} \delta q^j(t^{\prime}). \end{align} \tag{2"} $$ When we calculate the commutator $$\begin{align}\left[\frac{\delta}{\delta q^j(t^{\prime})}, \frac{d}{dt}\right] \stackrel{(1)+(2')}{=}& \sum_{m=0}^{\infty}\sum_{i=1}^n \left[\frac{\delta}{\delta q^j(t^{\prime})},q^{i(m+1)}(t)\right]\frac{\partial}{\partial q^{i(m)}(t)}\cr &-\sum_{\ell=0}^{\infty}\left[\frac{d}{dt}, \delta^{(\ell)}(t\!-\!t^{\prime})\right]\frac{\partial}{\partial q^{j(\ell)}(t)} \cr ~~=~~& \sum_{m=0}^{\infty} \delta^{(m+1)}(t\!-\!t^{\prime}) \frac{\partial}{\partial q^{i(m)}(t)}\cr &- \sum_{\ell=0}^{\infty} \delta^{(\ell+1)}(t\!-\!t^{\prime})\frac{\partial}{\partial q^{j(\ell)}(t)}\cr ~~=~~&0,\end{align} \tag{3} $$ we get zero!

2. If we extend ${\cal F}$ with functions of finite many other times, or with functionals (say, with internal time-integrations), or both, similar calculations show that the derivatives (1) & (2) commute.

OP is essentially asking the following.

Why the total time derivative $$ \frac{d}{dt}~=~\frac{\partial}{\partial t} + \sum_{m=0}^{\infty}\sum_{i=1}^n q^{i(m+1)}(t)\frac{\partial}{\partial q^{i(m)}(t)} \tag{1} $$ and the functional/variational derivative $$ \frac{\delta}{\delta q^j(t^{\prime})} \tag{2}$$ commute?

That's a good question. The intuitive reason is that the differentiations refer to different variables. But actual calculations make it less obvious (cf. e.g. eq. (3) below). Assume that the derivatives (1) & (2) act on some space ${\cal F}$.

1. For instance, say that ${\cal F}$ is the space of functions of the form $f(q^{i(m)}(t),t)$. Then we may write the functional derivative (2) as $$ \frac{\delta}{\delta q^j(t^{\prime})} ~=~ \sum_{\ell=0}^{\infty}\delta^{(\ell)}(t\!-\!t^{\prime})\frac{\partial}{\partial q^{j(\ell)}(t)}.\tag{2'}$$ When we calculate the commutator $$\begin{align}\left[\frac{\delta}{\delta q^j(t^{\prime})}, \frac{d}{dt}\right] &~=~ \sum_{m=0}^{\infty}\sum_{i=1}^n \left[\frac{\delta}{\delta q^j(t^{\prime})},q^{i(m+1)}(t)\right]\frac{\partial}{\partial q^{i(m)}(t)} -\sum_{\ell=0}^{\infty}\left[\frac{d}{dt}, \delta^{(\ell)}(t\!-\!t^{\prime})\right]\frac{\partial}{\partial q^{j(\ell)}(t)} \cr &~=~ \sum_{m=0}^{\infty} \delta^{(m+1)}(t\!-\!t^{\prime}) \frac{\partial}{\partial q^{i(m)}(t)} - \sum_{\ell=0}^{\infty} \delta^{(\ell+1)}(t\!-\!t^{\prime})\frac{\partial}{\partial q^{j(\ell)}(t)}~=~0\end{align} \tag{3} $$ we get zero!

2. If we extend ${\cal F}$ with functions of finite many other times, or with functionals (say, with internal time-integrations), or both, similar calculations show that the derivatives (1) & (2) commute.

OP is essentially asking the following.

Why the total time derivative $$ \frac{d}{dt}~=~\frac{\partial}{\partial t} + \sum_{m=0}^{\infty}\sum_{i=1}^n q^{i(m+1)}(t)\frac{\partial}{\partial q^{i(m)}(t)} \tag{1} $$ and the functional/variational derivative $$ \frac{\delta}{\delta q^j(t^{\prime})} \tag{2}$$ commute?

That's a good question. The intuitive reason is that the differentiations refer to different variables. But actual calculations make it less obvious (cf. e.g. eq. (3) below). Assume that the derivatives (1) & (2) act on some space ${\cal F}$.

1. For instance, say that ${\cal F}$ is the space of functions of the form $f(q^{i(m)}(t),t)$. Then we may write the functional derivative (2) as $$ \frac{\delta}{\delta q^j(t^{\prime})} ~=~ \sum_{\ell=0}^{\infty}\delta^{(\ell)}(t\!-\!t^{\prime})\frac{\partial}{\partial q^{j(\ell)}(t)},\tag{2'}$$ because an infinitesimal variation is of the form $$\begin{align}\int\! dt^{\prime} \sum_{j=1}^n \frac{\delta f(q^{i(m)}(t),t)}{\delta q^j(t^{\prime})}\delta q^j(t^{\prime}) &~=~\delta f(q^{i(m)}(t),t) ~=~\sum_{j=1}^n \sum_{\ell=0}^{\infty}\frac{\partial f(q^{i(m)}(t),t)}{\partial q^{j(\ell)}(t)}\delta q^{j(\ell)}(t)\cr &~=~\int\! dt^{\prime} \sum_{j=1}^n \sum_{\ell=0}^{\infty}\delta^{(\ell)}(t\!-\!t^{\prime})\frac{\partial f(q^{i(m)}(t),t)}{\partial q^{j(\ell)}(t)} \delta q^j(t^{\prime}). \end{align} \tag{2"} $$ When we calculate the commutator $$\begin{align}\left[\frac{\delta}{\delta q^j(t^{\prime})}, \frac{d}{dt}\right] &\stackrel{(1)+(2')}{=}~ \sum_{m=0}^{\infty}\sum_{i=1}^n \left[\frac{\delta}{\delta q^j(t^{\prime})},q^{i(m+1)}(t)\right]\frac{\partial}{\partial q^{i(m)}(t)} -\sum_{\ell=0}^{\infty}\left[\frac{d}{dt}, \delta^{(\ell)}(t\!-\!t^{\prime})\right]\frac{\partial}{\partial q^{j(\ell)}(t)} \cr &~~~=~~~ \sum_{m=0}^{\infty} \delta^{(m+1)}(t\!-\!t^{\prime}) \frac{\partial}{\partial q^{i(m)}(t)} - \sum_{\ell=0}^{\infty} \delta^{(\ell+1)}(t\!-\!t^{\prime})\frac{\partial}{\partial q^{j(\ell)}(t)}~=~0,\end{align} \tag{3} $$ we get zero!

2. If we extend ${\cal F}$ with functions of finite many other times, or with functionals (say, with internal time-integrations), or both, similar calculations show that the derivatives (1) & (2) commute.