Covariant and partial derivative commute?

Question

I know that we have for a function $\Gamma: (-\varepsilon,\varepsilon)^2 \rightarrow M$ we have (at least I think I know that this is true)

$$\nabla_{\frac{\partial \Gamma}{\partial s}} \frac{\partial \Gamma(s,t)}{\partial t} = \nabla_{\frac{\partial \Gamma}{\partial t}} \frac{\partial \Gamma(s,t)}{\partial s}.$$

I am not looking for a proof of this one!

Now I found a proof on page 2 of this reference (first line) click me where I was wondering what exactly happens between the second and third term.

I mean it looks pretty much like this, but I was wondering at which points you have to evaluate the function and in what way they commute with each other.

So there it says $$ \nabla_{\frac{\partial \Gamma}{\partial t}} \frac{\partial}{\partial s}|_{s=0} \Gamma = \nabla_{\frac{\partial \Gamma}{\partial s}} \frac{\partial}{\partial t}|_{s=0} \Gamma .$$

My problem is that the $t$ evaluation is completely missing here and I don't know whether I have to commute $t$ and $s$ evaluation, too. Thus, I don't know where the covariant derivative should be evaluated.

So if anybody could make this step more precise, this would totally answer my question.

Answer 1

In Riemannian Geometry book of Manfredo perdigão do carmo you can see one proof for your question in pag 68 in the symmetry lemma.

Sorry for my English.

you can see 1: https://i.stack.imgur.com/9DSDx.png

ps: The lemma 4.1 in image is the symmery lemma.