I'm gonna give you a different perspective. It might help you, as long as you're familiar with "little-o" notation.
First, let's rewrite a derivative as a local linear (affine to be pedantic) approximation. More on this here.
$$ x(t+h) = x(t)+h\,x'(t)+o(h) \tag1$$
As you can easily see, if you isolate $x'(t)$, you go back to the usual definition:
$$ x'(t)=\frac{x(t+h)-x(t)}{h}+o(1) $$
Notice that we're assuming $h$ arbitrarily small, as you do when you take the limit for $h\to0.$
You can do the same for partial derivatives:
\begin{align}
f(t+h,x)&=f(t,x)+h\,f_t(t,x)+o(h) \tag2\\
f(t,x+k)&=f(t,x)+k\,f_x(t,x)+o(k) \tag3
\end{align}
Now let's see what we can do for the total derivative:
\begin{align}
f\left(t+h,\,x(t+h)\right) &= f\left(t+h,\,x(t)+h\,x'(t)+o(h)\right) & &\text{(1)}\\
&=f(t+h,x)+k\,f_x(t+h,x)+o(k) & &\text{(3) with }k=h\,x'(t)+o(h)\\
&=f(t+h,x)+\left[h\,x'(t)+o(h)\right]f_x(t+h,x)+o(h) & &\text{in this case }o(k)=o(h)\\
&=f(t,x)+h\,f_t(t,x)+\left[h\,x'(t)+o(h)\right]f_x(t+h,x)+o(h) & &\text{(2)}
\end{align}
Notice that for ease of notation I wrote $x$ meaning $x(t)$. Since we're dealing only with first order derivative we can safely ignore higher order ones:
$$ f_x(t+h,x)=f_x(t,x)+h\frac{\partial}{\partial t}f_x(t,x)+o(h)=f_x(t,x)+o(1)$$
Thus
\begin{align}
f\left(t+h,\,x(t+h)\right) &=f(t,x)+h\,f_t(t,x)+h\,x'(t)\,f_x(t,x)+o(h)\\
&=f(t,x)+h\underbrace{\left[f_t(t,x)+x'(t)\,f_x(t,x)\right]}_{\frac{d}{dt}\,f\left(t,x(t)\right)}+o(h)
\end{align}
Hence
$$ \frac{d}{dt}\,f\left(t,x(t)\right) = f_t(t,x)+x'(t)\,f_x(t,x)$$
I know this isn't exactly what you asked for, but I prefer this notation rather than using the standard definition. I think this is more clear on what becomes a partial derivative and what vanishes because is infinitesimal of higher order.
