I'm reading section 6.3 of Manifold Theory by Daniel Martin, I don't understand the following proof. We have two coordinate systems: $\{x^i\}$ and $\{x^{i'}\}$, then the following equation relates the Christoffel symbols: $$ \Gamma_{j'k'}^{i'}= \frac{\partial x^j}{\partial x^{j'}}\frac{\partial x^k}{\partial x^{k'}}\frac{\partial x^{i'}}{\partial x^i}\Gamma_{jk}^i+\frac{\partial^2x^i}{\partial x^{j'}\partial x^{k'}}\frac{\partial x^{i'}}{\partial x^i}. $$ The proof is this: Proof of the equation.
There is only one step I don't understand. In the fifth equation I don't see why you can change the $k$ for an $i$ and make it a sum, I know that $\nabla_{\partial/\partial x^j}\frac{\partial}{\partial x^k}=\Gamma_{jk}^i\frac{\partial}{\partial x^i}$, which is a sum over $i$. But then what about the term $\frac{\partial^2x^k}{\partial x^j\partial x^{k'}}\frac{\partial}{\partial x^k}$? It is not a sum. Then how does it turn into $\frac{\partial^2x^i}{\partial x^j\partial x^{k'}}\frac{\partial}{\partial x^i}$, which is a sum over $i$. Any help is appreciated.
The term
$$\frac{\partial^2x^k}{\partial x^j\partial x^{k'}}\frac{\partial}{\partial x^k}$$
is a sum (using Einstein summation notation, we are summing over the index $k$). When you sum over an index, it doesn't matter if you change the name of the index, so
$$\frac{\partial^2x^k}{\partial x^j\partial x^{k'}}\frac{\partial}{\partial x^k} = \frac{\partial^2x^i}{\partial x^j\partial x^{k'}}\frac{\partial}{\partial x^i}.$$
Therefore
\begin{align*} &\ \frac{\partial x^j}{\partial x^{j'}}\left[\frac{\partial x^k}{\partial x^{k'}}\Gamma^i_{jk}\frac{\partial}{\partial x^i} + \frac{\partial^2x^k}{\partial x^j\partial x^{k'}}\frac{\partial}{\partial x^k}\right]\\ =&\ \frac{\partial x^j}{\partial x^{j'}}\left[\frac{\partial x^k}{\partial x^{k'}}\Gamma^i_{jk}\frac{\partial}{\partial x^i} + \frac{\partial^2x^i}{\partial x^j\partial x^{k'}}\frac{\partial}{\partial x^i}\right]\\ =&\ \frac{\partial x^j}{\partial x^{j'}}\left[\frac{\partial x^k}{\partial x^{k'}}\Gamma^i_{jk} + \frac{\partial^2x^i}{\partial x^j\partial x^{k'}}\right]\frac{\partial}{\partial x^i}. \end{align*}