Alright, here is an idea for simple rectangular path that have to be split up in their convex sub-rectangles.
To get these sub-rectangles I save three coordinates for later use, the most lower-left corner, the most upper-right as well as the concave corner. There are named after their sub-rectangle (a, b) and where they lie relative to this sub-rectangle (bl = bottom-left, tr = top-right) but, of course, they can named whatever you want (and TikZ allows). The top-right coordinate of the a sub-rectangle is found by using (@b@bl |- @b@tr), i.e. the upper-left corner of the b sub-rectangle.
Your coordinates on the path were very complex. I will use here (cumulative) relative coordinates but your original path can be used too, of course (see commented section).
I have filled the sub-rectangles for clarification.
The path picture provides a special rectangular node, the path picture bounding box that usually can accessed via (path picture bounding box) and their anchors with, say (path picture bounding box.south west). Though, I will use low-level PGF macros to extract the x and y values of the lower-left (south west) and the upper-right (north east) corner. At the and I’ll have the dimension of the path picture in the \pgf@?b dimensions.
The \pgftext is a very low-level macro to place text somewhere in the picture.
It will placed with its upper-right corner (right,top) at the north east anchor of the path picture (\pgf@x and \pgf@y still hold the values from the north east anchor).
The content of the \pgftext will be a \vrule of the path pictures height (\pgf@yb) with no depth and no width and a \vrule with the width of the path picture and no other dimensions. This essentially makes a (TeX) box the size of the path picture with no visible content.
This all is wrapped inside a two-argumenty style called hyper. But this is only an auxiliary style that is used by the hyperlink style. You can simply expand this with \href or whatever is needed.
The implementation in TeX/PGF makes it compatible and slightly faster.
With the help of the calc library you could do:
\path let \p{@ppbb@dim}=
($(path picture bounding box.north east)-(path picture bounding box.south west)$)
in node[inner sep=+0pt,
outer sep=+0pt,
anchor=north east,
at=(path picture bounding box.north east)]
{#1{#2}{\rule{\x{@ppbb@dim}}{0pt}\rule{0pt}{\y{@ppbb@dim}}}};
Code
\documentclass[10pt,letterpaper]{article}
\usepackage{tikz}
\usepackage[colorlinks=true,urlcolor=blue,filecolor=magenta]{hyperref}
\makeatletter
\tikzset{
hyper/.style 2 args={
path picture={%
\pgfpointanchor{path picture bounding box}{south west}%
\pgf@xb-\pgf@x
\pgf@yb-\pgf@y
\pgfpointanchor{path picture bounding box}{north east}%
\advance\pgf@xb\pgf@x
\advance\pgf@yb\pgf@y
\pgftext[at={\pgfqpoint{\pgf@x}{\pgf@y}},right,top]{#1{#2}{\vrule height\pgf@yb depth0ptwidth0pt\vrule height0ptdepth0ptwidth\pgf@xb}}%
}
},
hyperlink/.style={hyper=\hyperlink{#1}}
}
\makeatother
\begin{document}
\section{The Diagram}
\begin{center}
\begin{tikzpicture}
\node[draw,align=center] at (12, 0.2+.5) {Even \\ more};
\node[draw,align=center] at (12, 1.4+.5) {And \\ more1};
\node[draw,align=center] at (15, 2.8+.5) {Some \\ text};
\node[draw,align=center] at (15, 1.8+.5) {More \\ text};
%\draw[blue] (11,-.4+.5) coordinate (@a@bl) -- (13,-.4+.5)
% -- (13,1.3+.5) coordinate (@b@bl) -- (16.5,1.3+.5)
% -- (16.5,3.5+.5) coordinate (@b@tr) -- (11,3.5+.5) -| (@a@bl) -- cycle;
\draw[blue] (11,-.4+.5) coordinate (@a@bl)
-| ++ (3 ,1.7) coordinate (@b@bl)
-| ++ (3.5,2.2) coordinate (@b@tr) -| (@a@bl) -- cycle;
\fill[blue!20, fill opacity=.7, hyperlink=target] (@a@bl) rectangle (@b@bl |- @b@tr);
\fill[blue!40, fill opacity=.7, hyperlink=target] (@b@bl) rectangle (@b@tr);
\end{tikzpicture}
\end{center}
\section{This is the target}
\hypertarget{target}{Here is the target!}
\end{document}
Output

@<username>in your comment (as did I) so that the user gets notified.) I posted an answer that can be used for simple rectangular paths. For making the line itself hyper-active I could see a solution that uses decorations and places small boxes along the line, which would work with any arbitrary paths. – Qrrbrbirlbel Jun 24 '13 at 18:58