The best way to understand this phase shift is to solve and study solutions of the Helmholtz equation near the boundary between two dielectric mediums. You don't quite have to solve the full Maxwell equations: the assumption that the light field can be modelled by one scalar field (approximately equal to one transverse component of the electric field) rather than the $(\vec{E},\vec{H})$ vectors is known as scalar diffraction theory and justified in chapters 1 and 8 of Born and Wolf, "Principles of Optics".
First an intuitive explanation. When total internal reflexion happens, the field isn't abruptly turned around by the interface, it actually penetrates some distance beyond the interface as an evanescent field. The phenomenon is actually wholly analogous to quantum tunnelling by a first quantised particle field described by e.g. the Schrödinger or Dirac equation into regions which are, through their being at a higher potential than the particle's total energy, classically "forbidden" to the particle. Indeed, if you have a sandwich of lower refractive index material between two higher index materials such that an incoming wave is "totally internally reflected" from the first high-index to lower-index interface, then some of the light tunnels through the sandwich and again propagates freely (i.e. non evanescently) when it gets through the low refractive index layer. The power transmitted through the layer decreases exponentially with layer thickness, as with analogous quantum tunnelling through high but thin potential barrier problems.
So, given that the field penetrates some distance into the lower refractive index medium, the "effective" interface actually lies a small distance into the lower refractive index medium. The Goos-Hänchen phase shift is the phase delay arising from this short journey into and out of the lower index medium.
Now for some details. Let
$$\psi_i = \exp(i\,n_i \vec{k} \cdot \vec{r}) = \exp(i\,n_i(k_x x + k_y y))$$
be the incident field with the plane of polarisation in the $x-y$ plane with the $x$-axis being the interface and $n_i$ the refractive index for $y>0$. The point is now that the boundary sees a scalar field variation of $\exp(i\,n_i\,k_x\, x)$ such that $n_i\,k_x > n_t\,k$ where $k = 2\pi/\lambda = \sqrt{k_x^2+k_y^2}$ is the field's freespace wavenumber. So, to ensure continuity of the scalar field across the boundary, the $x$-component of the wavevector on the lower side of this boundary must also be $n_i\,k_x$. So what is the $y$-component of the wavevector in the lower medium. It has to be $k_y^\prime$ where $\sqrt{{k_y^\prime}^2 + (n_i k_x/n_t)^2} = k^2$ so as to fulfill the Helmholtz equation $(\nabla^2 +k^2 n_t^2)\psi = 0$ in the lower medium. Therefore, $k_y^\prime = \pm \sqrt{k^2-(n_i k_x/n_t)^2}$ which is imaginary by dint of the condition for total internal reflexion $n_i\, k_x>n_t\,k$. Now the solution $k_y = - \sqrt{k^2-(n_i k_x/n_t)^2}$ is unphsyical as it would have the field magnitude rising exponentially with penetration depth into the lower medium. So, in the lower medium, there is a field of the form:
$$\psi_{t,1}(x,y) = \exp(-\sqrt{(n_i\, k_x)^2 - (n_t\,k)^2}\,y + i\,n_i\,k_x\,x)$$
Fields that dwindle exponentially with distance into a medium like $\psi_{t,1}$ are known as evanescent fields (evanescere is classical Latin for "vanish"). In a fuller vector field analysis done by fully solving Maxwell's equations, one can work out the Poynting vector and show that such fields do not bear optical power with them. Instead, they are very like inductive and capacitive energy stores; they of course have an energy density but it shuttles back and forth between neighbouring regions in the medium and so the nett power flux through any surface over a whole period is nought.
There is also the reflected field:
$$\psi_r(x,y) = \gamma_r \exp(i\,n_i \vec{k}_r \cdot \vec{r}) = \gamma_r\,\exp(i\,n_i(k_x x - k_y y))$$
where $ \gamma_r$ is a yet-to-be found reflexion co-efficient. To uphold continuity of the scalar field at the boundary, this field also has a corresponding evanescent field $\psi_{t,2}(x,y) = \gamma_r \psi_{t,1}(x,y)$. How do we find $\gamma_r$; in scalar field theory it is chosen to make the normal derivative to the interface of the scalar field continuous across the interface. So we have:
$$(1+\gamma_r) \left.\partial_y \psi_t(x,y)\right|_{y=0} = \left.\partial_y\left(\gamma_r\,\exp(i\,n_i(k_x x - k_y y)) + \exp(i\,n_i(k_x x + k_y y))\right)\right|_{y=0}$$
so that:
$$-\frac{1+\gamma_r}{1-\gamma_r} = -i\,g$$
where
$$g=\frac{n_i\,k_y}{\sqrt{(n_i k_x)^2 - (n_t k)^2}}$$
and $\gamma_r$ is complex; on inverting the billinear relationship between $\gamma_r$ and $g$ we find that $\gamma_r$ is the unity magnitude complex number:
$$\gamma_r = -\frac{1 + i\,g}{1-i\,g}$$
i.e. to a phase delay of:
$$\phi_{GH}=-2\arctan\left(\frac{1}{g}\right) = -2\arctan\left(\frac{\sqrt{(n_i k_x)^2 - (n_t k)^2}}{n_i\,k_y}\right)$$
so its phase represents the Goos-Hänchen shift, a kind of "mean" radian penetration into the lower index medium by the field.
Best Answer
Reflection to the right in your diagram is called "transmission."
Your drawing doesn't show the phase of the light. Remember that light is an oscillating electromagnetic field subject to boundary conditions. At a dielectric boundary the electric field component parallel to the surface must be continuous, because there's no charge geometry that would explain a discontinuity; the electric field component perpendicular to the surface will be discontinuous, because there is electric field normal to the surface from bound surface charges in the polarized material. This difference between the parallel and perpendicular E-field boundary conditions is the reason we have Snell's Law in dielectrics.
You have light normal to the surface, so you must have $$ E_\text{incident} + E_\text{reflected} = E_\text{transmitted} $$ at both interfaces. However the material in the dielectric is polarized, with $$ (\epsilon-\epsilon_0)E = P $$ For the interface with internal reflection, the polarization $P$ contributes to both the incident and reflected waves. However for the interface with external reflection the polarization contributes only to the transmitted wave. I think this gives you a sign difference in the contribution of $P$ to the reflected wave, but I'm too sleepy to remember the details — I'm pretty sure it has something to do with a wave equation for $D=\epsilon E$.
The way you measure a phase difference, of course, is with an interference experiment.
Here's another way you can think of it. Your phase difference is with respect to the incident light; there'll be another phase difference because there's a path length difference between the internally- and externally-reflected light. You can reduce this path length difference by bringing your two surfaces closer together. In the limit where the two surfaces are separated by much less than the wavelength of light and are exactly parallel, the internally- and externally-reflected rays should be exactly out of phase and interfere destructively. But the limit where the two surfaces have zero separation is the same as the limit where there's no interface at all and the dielectric is continuous — in that case there is also no reflection, only transmission.