Why does altering path length in radio transmission alter the phase of the received signal?
Yes indeed it will be (although we could argue that it may not be necessarily for all distances since the phase is cyclical!). 

In free space the signal propagates at the speed of light, therefore this sets the wavelength in distance based on the frequency transmitted according to:

$$\lambda = c/f$$

Where $c$ is the speed of light in meters/second (or other units of choice), and $f$ is the frequency of the signal in units of  Hz if time is in units of seconds. A fixed tone at a given frequency with have a length in phase of $2\pi$ over the distance of one wavelength.

If the channel has a dielectric constant higher than free space given by $\epsilon_r$, the wavelength will be decreased by $1/\sqrt{\epsilon_r}$. 

For example the wavelength of a 1 GHz signal in free space (or air which has a dielectric constant that is very close to free space) is approximately $3e8/1e9 = 0.3$ meters, while over a cable with a polyethylene dielectric ($\epsilon_r = 2.3$) would be $3e8/(1e9\sqrt{2.3}) = 0.198$ meters. 

For this case of a 1 GHz signal in free space given a wavelength of $0.3 m$ (corresponding to the phase over one wavelength of $2\pi$ radians), if you receive the direct signal with no multi-path at two different locations with a $1$ cm path difference between the two, the expected phase difference will be $0.21$ radians.   



As a follow up to the OP's question; if you set $t=0$ upon arrival of the wavefront, then the phase everywhere at the new $t=0$ will be equal to the phase when the signal was initiated at the transmitter. Although this is somewhat of a meaningless result: it is equivalent to measurement the distance of two objects and resetting the distance to 0 once you determine the objects length.