Theorem 13.15: Let $\phi: G \rightarrow G'$ be a group homomorphism, $g \in G$.
Then $g\ker\phi = (\ker\phi)g = \operatorname{Im}^{-1} \left[ \; \{ \; \phi(g) \; \} \; \right] = \phi^{-1}[ \; \{ \; \phi(g) \; \} \;] = \{ \; x \in G : \phi(x) = \phi(g) \; \} \; (☼)$
Consequently, the two partitions of G into left cosets and into right cosets of H are the same.
I'm not asking about proofs.
(1.) How does this picture induce and flesh out the theorem? I don't understand either.
What's the intuition of the left and right cosets $g\ker\phi = (\ker\phi)g$?
(7.) I'm confounded by $(☼)$. How are all these complicated notations equal intuitively?
(2.) What do the dotted lines represent?
(3.) Why can Fraleigh represent 'the cosets of $\ker\phi$' by 'the solid vertical line segments'? How?
See that: $H=\text{ker }\phi$.
(1.) Don't forget that in non-abelian groups left and right cosets aren't the same unless the subset wich you are looking at is normal in $G$. The overall intuition on cosets is that, they partition the group into distinct subsets.
(2.) the dotted lines show the image of the set in the upper part of the picture under $\phi$.
(3.) See (1.). otherwise: it's just a picture, if this picture doesn't work for you, try to come up with another one, that suits you better. (wich would be a good exercise)
(4.) your statement is exactly the same. but see (5.)
(5.) actually: $\phi(y)=\phi(kg)=\phi(k)\phi(g)=\phi(g)$ for some $k\in \text{ker }\phi$
(6.) could you clarify your question? in the statement of the theorem it says: $g\in G$ ?! I don't get your point, sry.
EDIT: (6.) in the above theorem it says "$g\in G$" wich is short for "let $g\in G$". this is just the usual proof strategy: show the statement is true for some arbitrary element, hence is true for all, since you did not claim your element to satisfy any special property and this is true for all elements.