I'm not sure if this answers your question, but:
The dual of $X^*$ is $X^{**}$ and consists of the bounded linear linear functionals on $X^*$. Now, given a point $x$ in $X$, pointwise evaluation at $x$ by elements of $X^*$ defines an element of $X^{**}$. So, in a sense, you can view $x$ as an element of $X^{**}$. This is what $i_x$ does - $i_x$ is the element $x$ viewed as a linear functional on $X^*$.
In general, though, $X^{**}$ may contain other functionals. The weak topology on $X^*$ is induced by elements of $X^{**}$ while the weak* topology is induced by elements of $X$:
A basic nhood of $0$ in the weak topology of $X^*$ has the form $ \{ x\in X^* | |f_1(x)|<\epsilon, |f_2(x)|<\epsilon, \ldots, |f_n(x)<\epsilon\} $ for some $\epsilon>0$ and $f_1,\ldots, f_n$ in $X^{**}$
A basic nhood of $0$ in the weak$^*$ topology of $X^*$ has the form $ \{ x\in X^* | |f_1(x)|<\epsilon, |f_2(x)|<\epsilon, \ldots, |f_n(x)|<\epsilon\} $ for some $\epsilon>0$ and $f_1,\ldots f_n$ in $i(X)$
Note that the only difference between the two is that with the weak* topology, you are only using the functionals on $X^*$ defined by elements of $X$.
With regards to your third paragraph, when you speak of the weak topology on $X$, the map $i$ doesn't come into play.