Taylor expansion of Eulerian velocity field I struggle to understand the following taylor expansion:

I know this is nothing else than the "total differential" of the velocity field, but I encountered such Taylor expansion several times by now and I finally want to understand what is done here.
Can somebody please teach me how to expand $ \vec{\mathsf v}(t+\mathrm d t,\vec r+\mathrm d\vec r)$ so I obtain the total  differential $\mathrm d \vec{\mathsf v} \equiv \vec{\mathsf v}(t+\mathrm d t,\vec r+\mathrm d\vec r) - \vec{\mathsf v}(t,\vec r)$ as the lowest order approximation
Ok to clarify my issue: How do you expand $\vec{\mathsf v}(t+\mathrm d t, \vec r+\mathrm d \vec r)$ correctly as a „mathematician“. If one considers a function of more than one variable  $f: \mathbb{R}^n \rightarrow \mathbb R $, e.g. $f(x,y)$, the Taylor expansion reads $$f(x,y) \simeq f(a,b)+\left[(x-a)\partial_x+(y-b) \partial_y \right]f+\frac{1}{2!}\left[(x-a)\partial_x+(y-b)\partial_y\right]^2 f+ \dots$$ for $n=2$ at $(x,y) = (a,b)$. In order to expand $\vec{\mathsf v}(t+\mathrm d t, \vec r+\mathrm d \vec r)$ I would need to introduce variables $t'= t+\mathrm d t$. That is where my trouble starts.
I also thought about expanding $\vec{\mathsf v}(t, \vec r)$ around the point $(t+ \mathrm dt, \vec r + \mathrm d\vec r)$ but then I would not obtain  the desired result regarding the partial derivatives as they would also be evaluated at $(t+ \mathrm dt, \vec r + \mathrm d\vec r)$. Furthermore, the signs would turn from $+\rightarrow -$ in front of each term of $\mathrm d \vec{\mathsf v}$
Here is the algebra: where I tried to expand $\vec{\mathsf v}(t+\mathrm d t, \vec r+\mathrm d \vec r)$
Let $t' = t+\mathrm d t$ and $\vec{r'} = \vec{r}+\mathrm d \vec{r}$. The Taylor expansion up to the first order of $\vec{\mathsf{v}}\big(t' ,\vec{ r'}\big)$ at $ \big(t' ,\vec{ r'}\big) = (t, \vec r) $ fixed yields:
$$ \vec{\mathsf{v}}\big(t' ,\vec{ r'}\big) \simeq  \vec{\mathsf{v}}(t, \vec r) +
 \underbrace{\left(t'-t\right)}_{=\mathrm d t}\frac{\partial \vec{\mathsf{v}}\big(t' ,\vec{ r'}\big) }{\partial t'}\bigg|_{t, \vec r} + 
 \underbrace{\sum_{i=1}^{3}  \underbrace{\left({x^{i}}'-x^i\right)}_{=\mathrm d x^i}\frac{\partial \vec{\mathsf{v}}\big(t' ,\vec{ r'}\big) }{\partial {x^{i}}'}\bigg|_{t, \vec r}}_{\left(\mathrm d \vec r\cdot \vec{\nabla} ' \right)\vec{\mathsf{v}}(t', \vec r')  }$$
where $\vec r = \left(x^1, x^2, x^3\right)^{\mathsf T}$ and $\vec \nabla '= \left(\frac{\partial}{\partial{x^1}'}, \frac{\partial}{\partial{x^2}'}, \frac{\partial}{\partial{x^3}'}\right)^{\mathsf T} $
$$\Rightarrow \mathrm d \vec{\mathsf{v}} = \vec{\mathsf v} (t+\mathrm d t,\vec r+\mathrm d \vec r) - \vec{\mathsf v} (t,\vec r)  \simeq \frac{\partial \vec{\mathsf{v}}\big(t' ,\vec{ r'}\big) }{\partial t'}\bigg|_{t, \vec r} \mathrm dt + \left(\mathrm d\vec r\cdot \vec \nabla' \right) \vec{\mathsf{v}}(t', \vec r') \bigg|_{t, \vec r} $$
This is (at least to me) not the same as the desired expression $\mathrm d {\vec{\mathsf v}} \simeq \frac{\partial \vec{\mathsf v}(t, \vec r)}{\partial t} \mathrm d t + \sum_{i=1}^3 \frac{\partial \vec{\mathsf v}(t, \vec r)}{\partial x^i} \mathrm d x^i $ or can we assume that $$ \frac{\partial \vec{\mathsf{v}}\big(t' ,\vec{ r'}\big) }{\partial t'}\bigg|_{t, \vec r} \mathrm = \frac{\partial \vec{\mathsf{v}}\big(t ,\vec{ r}\big) }{\partial t}$$ (analog for the spatial derivatives) and if so, why?!
 A: To first order
$$\vec{v}(t+dt,\vec{r}+\vec{dr})=\vec{v}(t,\vec{r})+\dfrac{\partial \vec{v}}{\partial t}dt+ \nabla \vec{v}.\vec{dr}$$.
Notice how similar it is to:
$$f(x+dx) = f(x)+f'(x)dx$$
So this is really just multivariable Taylor expansion. Note that $\nabla \vec{v}$ is a matrix (or 2nd rank tensor, more properly) hence dotting it with a vector gives you back a vector. Also $\nabla$ refers only to the gradient with respect to the space variables ($\vec{r}$) in case that's not clear.
Edit:
Your 2 functions are indeed the same function with different names for the variables. If I have the function $f(x)=x$ I could just as well write it as $f(x')=x'$, it's the exact same function. So when I evaluate $f(x')$ at the point $x_0$ obviously it's going to be equal to $f(x_0)$.  Similarly we have that
$\dfrac{\partial \vec{v}(t',\vec{r}')}{\partial t'}$ defines exactly the same function as $\dfrac{\partial \vec{v}(t,\vec{r})}{\partial t}$ which in turn defines exactly the same function as $\dfrac{\partial \vec{v}(h,\vec{b})}{\partial h}$.
If I evaluate all of them at the point $(t,\vec{r})$ then all three of them will give me the same answer. I know it can be confusing since the quantity $$\dfrac{\partial \vec{v}(t,\vec{r})}{\partial t'}$$ may seem different from $$\dfrac{\partial \vec{v}(t,\vec{r})}{\partial t}$$
But in fact they're exactly the same since both of them should be read as: the time derivative of $\vec{v}$ evaluated at the point $(t,\vec{r})$.
If you're still not convinced and want a fully rigorous explanation you could do the following: Obtain the above two quantities as limits of difference quotients (which is what derivatives are) and observe that they're exactly the same.
