For a tangent vector $\bf v$ at a point $x\in\Gamma$, and a smooth (real valued) function $f$ defined on an open neighborhood of $x$, one can define
$\nabla_{\bf v}f:=\langle {\bf v},\nabla_\Gamma f\rangle$ (the scalar product).
Note that both $\bf v$ and $\nabla_\Gamma f$ already lie in the tangent space of $\Gamma$, and hence only the scalar product of $T_x\Gamma$ is used.
Then, for a vector field $Y$ (on $\Gamma$), if given by coordinate functions $(f_i)_{i=1}^n$, define $\nabla_{\bf v}Y := (\nabla_{\bf v} f_i)_i$.
Finally, for a point $x\in\Gamma$, set $(\nabla_X Y)_p := \nabla_{Xp}Y$.
To answer your stated questions: Yes, you've now done the computations correctly, and no, that's about as simple as you can have the expression. One may take advantage of the tensorial nature of the covariant derivative and write
$$ \nabla_i \omega_j = \partial_i \omega_j - \Gamma^k_{ij}\omega_k $$
using a bit of a sloppy mixture of abstract index notation with coordinate index notations, but that is about it.
To answer your question in the comments, we introduce the abstract notion of a derivation on the tangent bundle of a (smooth) manifold $M$. Let $\mathfrak{T}^{r,s}(M)$ denote the $\mathbb{R}$-vector space of smooth $(r,s)$ tensors over $M$. A tensor derivation is defined to be a $\mathbb{R}$-linear map $\mathscr{D}:\mathfrak{T}^{r,s}(M)\to \mathfrak{T}^{r,s}(M)$ defined for all pairs $(r,s)\in \mathbb{Z}_{\geq 0}^2$ satisfying the following conditions:
- If $A$ and $B$ are smooth $(r,s)$ and $(t,u)$ tensor fields respectively, we have that $\mathscr{D}(A\otimes B) = \mathscr{D}A\otimes B + A\otimes \mathscr{D}B$ (in other words, the Leibniz rule holds for tensor products).
- If $A$ is a smooth $(r,s)$ tensor field, and $\mathfrak{C}:\mathfrak{T}^{r,s}(M) \to \mathfrak{T}^{r-1,s-1}(M)$ is a tensor contraction, then we have $\mathscr{D}(\mathfrak{C}A) = \mathfrak{C}(\mathscr{D}A)$. (It commutes with tensor contractions.)
It is a theorem that for $r = s = 0$, any $\mathbb{R}$-linear map on $\mathfrak{T}^{0,0}(M)$ (the space of smooth functions) that satisfies the Leibniz rule (the second condition does not apply since there are no tensor contractions applicable to pure functions) can be represented as the directional derivative relative to a vector field. So we can add the unessential third condition
- There exists a smooth vector field $V$ such that for all smooth functions $f$ over $M$, $V(f) = \mathscr{D}f$.
Remark: This natural one-to-one correspondence between derivations and smooth vector fields allow us to also interpret a derivation as a map from $\mathfrak{T}^{1,0}(M)$, the space of smooth vector fields, into the space of $\mathbb{R}$-linear maps on smooth tensor fields that satisfies conditions 1 and 2. This manifests in the notation $\nabla_X Y$ for covariant differentiation and $\mathcal{L}_X Y$ for Lie differentiation.
In any case, just given the above three conditions are not enough to specify the derivation. But we just need one more ingredient: the action of $\mathscr{D}$ on either $\mathfrak{T}^{1,0}$ or $\mathfrak{T}^{0,1}$. I'll sketch the case of $\mathfrak{T}^{1,0}(M)$.
Suppose we know what $\mathscr{D}X$ is for all smooth vector fields. Then for one-forms $\omega$, we consider the contraction $\mathfrak{C}(\omega\otimes X)$ which we probably more commonly write as $\omega(X)$ and is a function. So using the axioms of the derivation, we have
$$ \mathscr{D}[\omega(X)] = (\mathscr{D}\omega)(X) + \omega(\mathscr{D}X) $$
this is an algebraic equation in which all objects are known except for one: we know what $\omega(X)$ is given $\omega$ and $X$, we know what $\mathscr{D}X$ is by assumption, and so also $\omega(\mathscr{D}X)$. We know that $\mathscr{D}[\omega(X)]$ is since $\omega(X)$ is a function. Hence we can solve the algebraic equation to get a formula for $\mathscr{D}\omega$, using that $\mathfrak{T}^{1,0}(M)$ and $\mathfrak{T}^{0,1}$ are duals so that to specify $\mathscr{D}\omega$ it suffices to specify $(\mathscr{D}\omega)(X)$ for all $X$.
Similarly, now given an arbitrary tensor field $\Xi\in \mathfrak{T}^{r,s}(M)$, we can do the same procedure and consider
$$ \mathscr{D}\left[ \mathfrak{C}_1\mathfrak{C}_2\cdots\mathfrak{C}_{r+s} \Xi \otimes X_1\otimes X_2\otimes\cdots\otimes X_s\otimes \omega_1\otimes\cdots\otimes \omega_r\right] $$
the full contraction of $\Xi$ with $s$ arbitrarily chosen vector fields and $r$ arbitrarily chosen one forms. Expanding the above expression using the axioms of the derivation, we are left with only one unknown: $\mathscr{D}\Xi$. Everything else $\mathscr{D}X_n$, $\mathscr{D}\omega_n$ etc are all computable from the axioms, the vector field $V$ (of the third axiom), and the assumed knowledge of $\mathscr{D}X$ for all vector fields $X$.
For illustration, we can verify that the above construction is indeed "tensorial". Given that
$$ \mathscr{D}\omega(X) = \mathscr{D}(\omega\cdot X) - \omega(\mathscr{D}X) $$
one may wonder whether $\mathscr{D}\omega$ is indeed a one-form: that is, whether $\mathscr{D}\omega(fX) = f\mathscr{D}\omega(X)$ for $f$ a smooth function. We can directly compute:
$$ \mathscr{D}\omega(fX) = \mathscr{D}(\omega\cdot fX) - \omega(\mathscr{D}(fX)) = \mathscr{D}[f(\omega\cdot X)] - \omega\left( \mathscr{D}f X + f \mathscr{D}X\right) = \mathscr{D}f (\omega\cdot X) + f \mathscr{D}(\omega\cdot X) - f \omega(\mathscr{D}X) - \omega(\mathscr{D}f X) $$
Noting that $\mathscr{D}f(\omega\cdot X) = \omega(\mathscr{D}f X)$ by the tensorial property of $\omega$ as a one form, we see that indeed the Leibniz-rule + contraction rule allows us to make sure that $\mathscr{D}\omega$ (and hence $\mathscr{D}\Xi$ for an arbitrary tensor field $\Xi$) is tensorial.
Lastly, what does this have to do with connection coefficients? All this mucking about with the Christoffel symbols is just a way of saying: we know what $\mathscr{D}X$ is. That is, consider the derivation labeled $\nabla_Y$, which acts on scalar fields as the vector field $Y$ with coordinate expansion $Y^i\partial_i$. For an arbitrary vector field $X$ given in local coordinates $X^i \partial_i$, we demand that the local coordinate expression of $(\nabla_Y X)^i \partial_i$ be given by
$$ (\nabla_Y X)^i = Y^j\partial_j X^i + \Gamma^i_{jk}X^k Y^j $$
and we carry on from there.
Best Answer
If you mean the covariant derivative of $\mathbb R^3$, then yes, but since the covariant derivative coincides with the Euclidean derivative it's pretty much just rewriting the formula you already have as $$ {\bf s}_t = {\bf \nabla_n u - n\cdot(\nabla_n u)\;n}.$$
If you mean the covariant derivative of $\Gamma$, the answer is no: ${\bf s}_t$ depends explicitly upon the normal derivative while the covariant derivative of $\Gamma$ only measures changes in tangential directions.
If you define the "$\Gamma$-tangential derivative" of $u$ by $\nabla^\Gamma_v u = v\cdot\nabla u - ({\bf n} \cdot v\cdot \nabla u){\bf n}$ then it's true that ${\bf s}_t = \nabla_{\bf n} \bf u$ and it's also true that $\nabla^\Gamma$ agrees with the covariant derivative where the latter is defined. The issue is that $\bf n$ is not tangent to $\Gamma$, so it's not correct to call $\nabla^\Gamma_\bf nu$ a covariant derivative.