Gaussian connection on submanifold in Euclidean space: how to compute Christoffel coefficients Consider an embedding
$$\boldsymbol{r}=\boldsymbol{r}(u^1,\ldots,u^n)$$
of some $n$-dimensional manifold $M$ in $\mathbb{R}^N,$
with the correspondind induced metric.
We know that both $\boldsymbol{r}_i=\frac{\partial \boldsymbol{r}}{\partial u^i}$ and $\frac{\partial}{\partial u^i}$ are a basis for $T_uM;$ consider now two vector fields $X,Y$ in $\mathbb{R}^N$ tangent to $M$ and define the Gaussian connection
$$ \left. \nabla_X Y \right|_{u\in M}:= \operatorname{pr}_u \partial_XY. $$
My question is: Why is the following computation possible, namely, why do we interchange the two basis of tangent space as if they are the same (I know they are morally the same, but I'm lacking a formal justification of this fact) ?
$$ \nabla_{\partial/\partial u^i} \frac{\partial}{\partial u^j}= \operatorname{pr}_u \frac{\partial}{\partial u^i} \boldsymbol{r}_j.$$
 A: Your final equation only holds because you are using the flat connection on $\mathbb{R}^N$. 
Let us generalize the picture a little bit to see more of the structure. 
Consider $M,N$ manifolds, and let $g$ be a Riemannian metric on $N$. Let $\phi:M\to N$ be an embedding. Then $\phi^*g$ is the induced/pullback metric on $M$. 
Let $u_i$ be a local coordinate system on $M$. Let $X,Y$ be vector fields on $M$. What we have for the induced connection $\nabla$ (the Levi-Civita connection relative to $\phi^*g$) is that the pushforward 
$$ \phi_*\left(\nabla_X Y\right) = \operatorname{pr}_{T\phi(M)} D_{\phi_*X}\phi_*Y $$
where $\operatorname{pr}_{T\phi(M)}$ indicate the orthogonal projection to the tangent space $T\phi(M)$ under the metric $g$ of a vector in $TN$. $D$ is the Levi-Civita connection of the metric $g$ on $N$. 
Now, in general
$$ D_{\phi_* \partial/\partial u^i} \phi_* \partial/\partial u^j \neq \frac{\partial^2 \phi}{\partial u^i \partial u^j} $$
The two sides do not even live in the same space!
Note that $\phi: M \to N$ we have that $d\phi : TM\to TN$ so that $\partial_i \phi: M \to TN$ has co-domain the tangent bundle $TN$. Its derivative
$$ d(\partial_i \phi): TM \to T(TN) $$
means that the object $ \partial^2_{ij}\phi$ on the right hand side lives in the second tangent space $T(TN)$, and cannot be identified the left-hand side which is an element of $TN$. 
The Euclidean connection $\partial$, which is the Levi-Civita connection associated to the Euclidean metric $e$ on $\mathbb{R}^N$, however, has the following nice property: let $T\mathbb{R}^N = \mathbb{R}^N\times \mathbb{R}^N$ be the canonical splitting based on the standard coordinates of $\mathbb{R}^N$. Let $\pi_2: T\mathbb{R}^N \to \mathbb{R}^N$ be the projection to the second factor. Let $V,W$ be vector fields (sections of $T\mathbb{R}^N$). We have that
$$ \partial_V W = V\cdot d(\pi_2 W) $$
where $\pi_2 W: \mathbb{R}^N\to\mathbb{R}^N$ is the way one thinks about a vector field in analytic geometry, that is, as a mapping from $\mathbb{R}^N$ to itself. 
It is through this wonderful coincidence that you can write your expression to have the right hand side as a Hessian of $\mathbf{r}$. 

All said and done, though, perhaps your question is at a more elementary level. In which case maybe I can point out that by the chain-rule, if $\mathbf{r}:M\to\mathbb{R}^N$ and $f: \mathbb{R}^N \to V$, then 
$$ \frac{\partial}{\partial u^i} (f\circ \mathbf{r}) = (\partial_{\mathbf{r}_i} f)\circ \mathbf{r} $$
Hence if we think about the vector field $\mathbf{r}_j$ as a $\mathbb{R}^N$ valued function defined along $\mathbf{r}(M) \subset \mathbb{R}^N$, we have that $\frac{\partial}{\partial u^i} \mathbf{r}_j = \partial_{\mathbf{r}_i} \mathbf{r}_j$ (where by abuse of notation we omit certain compositions with $\mathbf{r}$). 
