I'm following the lecture here, and this question is with respect to the content on the whiteboard at the timestamp provided.
Consider Newtonian spacetime $(\mathcal{M}, \mathcal{O}, \mathcal{A}, \nabla, t)$ where $(\mathcal{M}, \mathcal{O}, \mathcal{A})$ is a smooth 4-dimensional manifold where all charts $(\mathcal{U}, x)$ are of the form $$x^0 : \mathcal{U} \to \mathbb{R}\\ x^1 : \mathcal{U} \to \mathbb{R}\\ \vdots\\ x^3 : \mathcal{U} \to \mathbb{R} $$ where $x^0 = t|_\mathcal{U}$ is the restriction of the absolute time function $t$ to the chart domain $\mathcal{U}$ and $\nabla$ is a prescribed covariant derivative operator with given connection coeffitient functions.
I'm trying to reevaluate the claim on the right, namely $$ 0 = \nabla dt.$$ As I understand it, one does so by picking any direction $\frac{\partial}{\partial x^a}$ from a chart-induced basis for $a=0,1, \ldots, 3$, then act on this vectorfield with the prescribed covariant derivative operator and apply the result to the $(0,1)$-tensor field $dt$, which is the gradient of the 0th component function: $d(x^0)$. The action of the covariant derivative along a vectorfield on a $(p, q)$-tensor field yields again a $(p, q)$-tensor field, hence we can look at the resulting $(0,1)$-tensor-field, i.e. a vector field, component-wise for components $b=0, \ldots, 3$. Formally, we apply the rules for covariant derivatives $$ \left(\nabla_{\frac{\partial}{\partial x^a}} d(x^0)\right)_b = \frac{\partial}{\partial x^a}\left(d(x^0)_b\right) - \Gamma_{b m}^n d(x^0)_n \left(\frac{\partial}{\partial x^a}\right)^m $$ According to the notes on the blackboard, the first term should vanish entirely, while the second one should reduce to $\Gamma_{ba}^0$. I'm havin trouble to see this.
Let's start with the first term: $d(x^0)$ is a covector, so $d(x^0)_b$ is the 0th component function. $\left(\frac{\partial}{\partial x^a}\right)$ is a vector field, which can be applied to a fucntion to yield another function. Alright. Still I can't see how I need to proceed there.
Secondly, I'm stuck with the rightmost term. Somehow, $\left(\frac{\partial}{\partial x^a}\right)^m$ must reduce to $\delta_a^m$ and $d(x^0)_n$ must reduce to $\delta_n^0$, however I don't see how.
Basically, as far as I understand, the whole expression needs to be the zero function in the end.
Can anyone explain where I seem to go wrong? Or help me understand the step missing?
Edit 1
After the commetnt by @willie-wong, I think I can answer the first part. First I get the component functions of $dx^0$ by means of $$ (df)_j := (df) \left(\frac{\partial}{\partial x^j}\right) = \left(\frac{\partial}{\partial x^j}\right)(f).$$ In the case I presented this would result in $$(dx^0)_b = \frac{\partial x^0}{\partial x^b} =\delta_b^0, $$ which results in the colletion of components $(1, 0, 0, 0)$ and $$ \frac{\partial}{\partial x^a}\delta_b^0 = \partial_a(\delta_b^0 \circ x^{-1}) = 0, $$ since the output of $\delta$ is constantly 0 or one, depending on $b$ with respect to any index $a$. Makes sense to me.
Edit 2
As in edit 1, the term $d(x^0)_n$ becomes $\delta_n^0$ following the same calculation. In regards to the second term, I may interpret $$ \left( \frac{\partial}{\partial x^a}\right)^m = \left( \frac{\partial x^m}{\partial x^a}\right),$$ i.e. as the $a$-th basis vector applied to the $m$-th component function. Hence the result doubtlessly is the $\delta_a^m$ function, and hence the second term is nonvanishing only for $n=0$ and $m=a$, resulting in the "time flows uniformly"-equation presented by Prof. Schuller.