How did Ramanujan get from $(1.1)$ to $(1.2)$?
Well, I can't promise this is how Ramanujan personally came up with it, but here's a derivation:
\begin{align}
\color{red}{\sum_{k=1}^{3n+1}\left(\frac 1{n+k}\right)}-\color{blue}{\sum_{k=1}^n\left(\frac 1{2n+2k}\right)} &= \color{red}{\left[\sum_{k=1}^{n}\left(\frac 1{n+k}\right) + \sum_{k=n+1}^{3n+1}\left(\frac 1{n+k}\right)\right]}-\color{blue}{\sum_{k=1}^n\left(\frac 1{2n+2k}\right)}\,\textrm{split red sum into pieces}\\
&=\sum_{k = 1}^n\left( \color{red}{\frac 1{n+k}} - \color{blue}{\frac 1{2n+2k}}\right)+ \color{red}{\sum_{k = n+1}^{3n + 1}\left(\frac 1{n+k}\right)}\\
&\textrm{pair first red sum and blue sum because the values of }k\textrm{ that they're summing over are the same}\\
&= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+ \color{red}{\sum_{k = n+1}^{3n + 1}\left(\frac 1{n+k}\right)}\,\textrm{red and blue combine to green}\\
&= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+\color{red}{\sum_{j = 0}^{2n}\left(\frac 1{n+j + n + 1}\right)}\\
&\textrm{reindex red sum as sum over }j\textrm{ by substituting }k = j + n + 1\\
&= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+\color{red}{\sum_{k = 0}^{2n}\left(\frac 1{2n+k+1}\right)}\\
&\textrm{the name of the dummy index doesn't matter, so change }j\textrm{ to }k\textrm{ in red sum}\\
&= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+ \left[\color{red}{\sum_{\substack{0\leq k\leq 2n\\k\textrm{ even}}}\left(\frac 1{2n+k + 1}\right)} + \color{blue}{\sum_{\substack{0\leq k\leq 2n\\k\textrm{ odd}}}\left(\frac 1{2n+k + 1}\right)}\right]\\
&\textrm{split up red sum into the sum over even }k\textrm{ (red) plus the sum over odd }k\textrm{ (blue)}\\
&= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+ \left[\color{red}{\sum_{k = 0}^n\left(\frac 1{2n+(2k)+ 1}\right)} + \color{blue}{\sum_{k = 1}^{n}\left(\frac 1{2n+ (2k - 1) + 1}\right)}\right]\\
&\textrm{even numbers are those of the form }2i\textrm{ and odd numbers are those of the form }2i - 1.\\
&\textrm{Use these to index even and odd sums explicitly}\\
&= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+ \left[\color{red}{\sum_{k = 0}^n\left(\frac 1{2n+2k+ 1}\right)} + \color{blue}{\sum_{k = 1}^{n}\left(\frac 1{2n+ 2k }\right)}\right]\,\textrm{simplify}\\
&= (\color{green}{1} + \color{blue}{1})\sum_{k = 1}^n\left(\frac 1{2n+2k}\right) + \color{red}{\sum_{k = 0}^n\left(\frac 1{2n+2k+ 1}\right)}\\
&\textrm{green and blue sums are the same, so combine them}\\
&= \sum_{k = 1}^n\left(\frac 2{2n+2k}\right)+ \color{red}{\sum_{k = 0}^n\left(\frac 1{2n+2k+ 1}\right)}\\
&2(a_1 + a_2 + \dots + a_n) = (2a_1 + 2a_2 + \dots + 2a_n)\textrm{ (i.e., move the }2\textrm{ to the inside of the sum)}\\
&= \sum_{k=1}^n\left(\frac 1{n+k}\right)+\color{red}{\sum_{k=0}^n\left(\frac 1{2n+2k+1}\right)}\,\textrm{simplify}.
\end{align}
How do you get the second set of equations?
\begin{align*}
\sum\limits_{k=1}^{4n+1}\frac 1k-\frac 12\sum\limits_{k=1}^{2n}\frac 1k-\frac 12\sum\limits_{k=1}^n\frac 1k &= \sum\limits_{k=1}^{4n+1}\frac 1k-\sum\limits_{k=1}^{2n}\frac{1}{2k}-\sum\limits_{k=1}^n\frac{1}{2k} \\
&= \sum\limits_{k=1}^{4n+1}\frac{1}{k} + (1 - 2)\sum\limits_{k=1}^{2n}\frac{1}{2k}-\sum\limits_{k=1}^n\frac{1}{2k}\\
&= \sum\limits_{k=1}^{4n+1}\frac{1}{k} + \sum\limits_{k=1}^{2n}\frac{1}{2k} - 2\sum\limits_{k=1}^{2n}\frac{1}{2k} -\sum\limits_{k=1}^n\frac{1}{2k}\\
&= \sum\limits_{k=1}^{4n+1}\frac{1}{k} + \frac{1}{2}\sum\limits_{k=1}^{2n}\frac{1}{k} - 2\sum\limits_{k=1}^{2n}\frac{1}{2k} -\sum\limits_{k=1}^n\frac{1}{2k}\\
&= \sum\limits_{k=1}^{4n+1}\frac 1k-2\sum\limits_{k=1}^{2n}\frac 1{2k}+\frac 12\sum\limits_{k=1}^{2n}\frac 1k-\sum\limits_{k=1}^n\frac 1{2k}.
\end{align*}
Now if you write out each of these four sums, you get
$$
\underbrace{\left[\frac{1}{1} + \dots + \frac{1}{4n + 1}\right]}_{\textrm{all denominators}} - 2\underbrace{\left[\frac{1}{2} + \dots + \frac{1}{4n}\right]}_{\textrm{even denominators}} + \frac{1}{2}\underbrace{\left[\frac{1}{1} + \dots + \frac{1}{2n}\right]}_{\textrm{all denominators}} - \underbrace{\left[\frac{1}{2} + \dots + \frac{1}{2n}\right]}_{\textrm{even denominators}},
$$
which can be viewed as
$$
\underbrace{\left[\frac{1}{1} + \dots + \frac{1}{4n + 1}\right]}_{\textrm{all denominators}} - 2\underbrace{\left[\frac{1}{2} + \dots + \frac{1}{4n}\right]}_{\textrm{even denominators}} + \frac{1}{2}\left(\underbrace{\left[\frac{1}{1} + \dots + \frac{1}{2n}\right]}_{\textrm{all denominators}} - 2\cdot\underbrace{\left[\frac{1}{2} + \dots + \frac{1}{2n}\right]}_{\textrm{even denominators}}\right).
$$
Now you have a $1/k$ for each $1\leq k\leq 4n + 1$ in the first sum, and in the second sum you take away $2/k$ for each even $k$ between $1$ and $4n + 1$. This means that if you combine these, you wind up with $1/k$ for $1\leq k\leq 4n + 1$ when $k$ is even, and $-1/k$ for $1\leq k\leq 4n + 1$ when $k$ is odd. The same logic applies to the last two sums, and you wind up with the desired
$$
\sum\limits_{k=1}^{4n+1}\frac {(-1)^{k+1}}k+\frac 12\sum\limits_{k=1}^{2n}\frac {(-1)^{k+1}}k.
$$
(P.S. you have an extra $k$ floating around in your problem statement - it should be $\sum\limits_{k=1}^{4n+1}\frac {(-1)^{k+1}}k+\frac 12\sum\limits_{k=1}^{2n}\frac {(-1)^{k+1}}k$, not $\sum\limits_{k=1}^{4n+1}\frac {(-1)^{k+1}}k+\frac 12\sum\limits_{k=1}^{2n}\frac {(-1)^{k+1}}k k$.)
What was Ramanujan's thinking when he split$$-\frac 12\sum\limits_{k=1}^{2n}\frac 1{k}=-2\sum\limits_{k=1}^{2n}\frac 1{2k}+\frac 12\sum\limits_{k=1}^{2n}\frac 1k$$
Again, I can't explain Ramanujan's personal thinking. He was extremely clever, and probably saw all the calculations performed above at a glance, and recognized that splitting up the sum in that way would let him combine terms into a pleasing/useful expression.