The concrete setting: Let $M\subset \mathbb{R}^3$ be a regular surface. A vector field $X$ in $M$ is a differentiable function $X: M\to \mathbb{R}^3$ such that $X(p)\in T_pM$.
Here we are taking the following "immersed" definition of $T_pM$: it is the image of $d\varphi_0$, where $\varphi:U\subset \mathbb{R}^2\to M$ is a diffeomorphism onto its image, such that $\varphi(0)=p$.
Then the standard connection $\nabla$ in $M$ is defined as $\nabla_X Y(p)=P_{T_pM}(dY_p(X(p)))$
where $P_{T_pM}:\mathbb{R}^3\to T_pM$ is the orthogonal projection.
The problem: I'm trying to verify that this connection is symmetric, but I'm having a clash of definitions.
On one hand, $$(\nabla_X Y-\nabla_y X)(p)=P_{T_pM}(dY_p(X(p))-dX_p(Y(p)))\hspace{2cm} \mbox{(1)}$$
Now I'd like to calculate the Lie bracket, and here is where I run into trouble.
The abstract setting: The definition of the Lie bracket I have is for two vector fields on an abstract manifold, where the tangent space at $p$ is the vector space of derivations at $p$. In this setting, a vector field is a section of the tangent bundle. If $X,Y$ are vector fields in $M$ and $f\in C^\infty(M)$, we have: $$[X,Y](p)(f):= X(p)(Yf) - Y(p)(Xf)$$
where $(Xf)(p):= (X(p))(f)=df_p(X(p))(\operatorname{id}_{\mathbb{R}})$.
Thus, $$[X,Y](p)(f)=d(Yf)_p(X(p))(\operatorname{id}_\mathbb{R}) - d(Xf)_p(Y(p))(\operatorname{id}_\mathbb{R}) \hspace{2cm} \mbox{(2)}$$
Back to the problem: I should prove that (1) and (2) are... "equal", but the concrete setting vs. the abstract setting clash: (1) is a vector in $T_pM$ while (2) is a real number.
I think I should reinterpret one in terms of the other, but I'm failing to do so. If I could interpret (1) in the abstract setting, maybe when I evaluate it at a function $f$ I would get (2). But I don't know how to properly justify it.