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Choosing a system of coordinates $(x_1,\ldots,x_n)$ about $p$ and writing $$ X = \sum_i x_i X_i, \qquad Y = \sum_j y_j X_j $$ where $X_i = \frac{\partial}{\partial x_i}$, we have \begin{align} \nabla_X Y &= \sum_i x_i \nabla_{X_i} \left(\sum_j y_j X_j \right) \\ &= \sum_{ij} x_i y_j \nabla_{X_i} X_j + \sum_{ij} x_i X_i(y_j)X_j \\ &= \sum_k \left(\sum_{ij} x_i y_j \Gamma_{ij}^k + X(y_k) \right) X_k, \end{align} the last inequality being justified by setting $\nabla_{X_i} X_j = \sum_k \Gamma_{ij}^k X_k$, where $\Gamma_{ij}^k$ are differentiable functions.

(from Manfredo Perdigão do Carmo, Riemannian geometry, pp. 50-51)

Question: I see clearly that setting $\nabla_{X_i} X_j = \sum_k \Gamma_{ij}^k X_k$ gives way to the first term of the last equality, but how is it also true that $$ \sum_{ij} x_i X_i(y_j)X_j = \sum_k X(y_k) X_k, $$ so that the last equality follows?

Edit: I figured it out. Indeed, it was because $$ \sum_{ij} x_i X_i(y_j)X_j = \sum_{j} X(y_j)X_j = \sum_k X(y_k) X_k. $$

N.B.: My background in differential geometry and tensor analysis is basic, so apologies for this basic question. I am hoping the answers can be stated with as much detail as possible.

Also, I think the symbol $\Gamma_{ij}^k$ denotes the Christoffel symbol, which to my understanding is very useful for finding the covariant derivative.

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    When you apply the connection to a product fX of a function f and a vector field X you get two terms... – 2017-01-21
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    In the case of do Carmo's textbook, are you possibly referring to the function $x_i$ and the vector field $X_i(y_j)$? – 2017-01-21
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    When you apply $\nabla_X$ to a product $fY$ of a function $f$ and a vector field $X$, you get a sum of two terms. That is what I am saying. What are those two terms? – 2017-01-21
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    $X(f) Y+f \nabla_X Y$? ... I guessed on the first term... – 2017-01-21
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    Don't guess. This is part of the definition of what an affine connection is. – 2017-01-21
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    Wait, that is, according to do Carmo's book, property (iii) of the definition an *affine connection*: $\nabla_X (fY)=f\nabla_X Y + X(f)Y$. Property (i) is $\nabla_{fX+gY} Z=f\nabla_X Z + g\nabla_YZ$ and property (ii) is $\nabla_X(Y+Z)=\nabla_X Y + \nabla_X Z$. This connection seems to behave like a linear operator so far. – 2017-01-21
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    If Do Carmo say that $\nabla_X (fY)=f\nabla_X Y + \nabla_X Z$ in his part (iii) of the definition, he is wrong. In fact, that does not even make sense, as what is Z? – 2017-01-21
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    No, Do Carmo did not say that; I typed it wrong. And I edited my comment accordingly, please refresh this page. – 2017-01-21
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    If you know that, then honestly I cannott see what you are asking. What is $\nabla_{X_i}(y_jX_j)$? – 2017-01-21
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    $\nabla_{X_i}(y_jX_j) = y_j \nabla_{X_i}X_j + X_i(y_j)X_j$ ... using property (iii) with $f=y_j$, $Y=X_j$, – 2017-01-21
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    Well, that 's where the second equality in the text you quoted comes from, precisely. The third one is just taking a common factor. – 2017-01-21
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    The common factor that we take is $\sum_k X_k$, right? This is actually my original question that I was asking, and I edited my question to be more explicit. – 2017-01-21
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    the common factor is $X_k$. – 2017-01-21
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    In order to take the common factor of $X_k$, I think I have to first show that $ \sum_{ij} x_i X_i(y_j)X_j = \sum_k X(y_k) X_k, $. – 2017-01-21
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    You just need to use the fact that $\displaystyle\sum_{i=1}^na_ib_i+\sum_{i=1}^na_i'b_i=\sum_{i=1}^n(a_i+a_i')b_i$, which is simply the distributivity of products over sumss. – 2017-01-21
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    And I need to use also that $X = \sum_i x_i X_i$ as well (which I mentioned at the very top of my question), so that $X(y_j)=\sum_j x_j X_j(y_j)$. (I figured that out just now.) Hence, $$\sum_{ij} x_i X_i(y_j)X_j = \sum_i X(y_j)X_j = \sum_k X(y_k)X_k.$$ Then to obtain the final expression, use what you just wrote, and pull out the common factor $\sum_k X_k$ (I still don't think the common factor is just "$X_k$".). I hope that is correct; I suddenly feel confident now. – 2017-01-21

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Note: do Carmo overuses the symbols $x_1,\dots,x_n$. They are the coordinates around $p$ and they are also the coefficients of the vector field $X$ in the induced basis $X_1,\dots,X_n$: $$X = \sum_i x_i X_i.$$ This is misleading, so it is preferable to replace the vector field $X$ with $Z$, and so we compute $\nabla_Z Y$, where $$ Z = \sum_i z_i X_i, \qquad Y = \sum_j y_j X_j. $$ Here, $z_i, y_j$ are smooth functions on the coordinate neighbourhood of $p$ that we are working with.

Now, we wish to know how the equality $$ \sum_{ij} z_i X_i(y_j) X_j = \sum_k Z(y_k) X_k \tag{$*$} $$ arises. This happens because $$ \sum_{ij} z_i X_i(y_j) X_j = \sum_j \left( \sum_i z_i X_i(y_j) \right) X_j,\tag{1} $$ and $$ Z(y_j) = \left( \sum_i z_i X_i \right)(y_j) = \sum_i z_i X_i(y_j).\tag{2} $$ Now, we just substitute $(2)$ back in $(1)$, and change the dummy variable $j$ to $k$, to get the equality $(*)$.