Physical reason for Lorentz Transformation Seeing the mathematical derivation of the Lorentz Transformation for time coordinates of an event for two observers we get the term
$$t'=-\frac{v/c^2}{\sqrt{1-\frac{v^2}{c^2}}}x+\frac1{\sqrt{1-\frac{v^2}{c^2}}}t$$
Now how to make sense physically of the $t-\frac{vx}{c^2}$ factor.
I am looking for an argument along the lines of the following. When relating spatial coordinates, one observer measures the length separation between an event and the second observer in his frame and tells the other observer that this should be your length, which the second observer denies due to relativity of simultaneity and multiplies by the $\gamma$ factor to get the correct length.

The physical reason IS the constancy of the velocity of light... since I'm writing in a tablet the answer won't be complete, but expect to get you to the mathematical cross-road.
Constancy of velocity of light implies that
\begin{equation}
  \frac{d|\vec{x}|}{dt} = c, \quad\Rightarrow\quad d|\vec{x}| = c\,dt.
\end{equation}
Since $d|\vec{x}| = \sqrt{dx^2 + dy^2 + dz^2}$, it follows that
\begin{equation}
  dx^2 + dy^2 + dz^2 = c^2\,dt^2 \quad\Rightarrow\quad dx^2 + dy^2 + dz^2 - c^2\,dt^2 = 0.
\end{equation}
From here it is straightforward to see that the set (or group) of transformations preserving this quantity are those known as Lorentz transformations.
Now I leave you to analize the "generalization" to nonvanishing intervals, for massive particles. Hint: define a four-dimensional metric!

(continuation... after a few days)
The interval
As exposed previously, the physical condition of constancy of the speed of light leads to the conclusion that 

All equivalent observers are connected through a transformation which
  keep the quantity $$dx^2 + dy^2 + dz^2 - c^2\,dt^2 = 0.$$

This can be generalized to the preservation of the quantity $$ I = dx^2 + dy^2 + dz^2 - c^2\,dt^2, $$ called interval.
Notice that the interval can be written as 
$$
I = X^t\, \eta\, X =
\begin{pmatrix}
ct & x & y & z
\end{pmatrix}
\begin{pmatrix}
-1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
ct \\
 x \\
 y \\
 z 
\end{pmatrix}
$$
Invariance of the interval
In order for the interval to be invariant under a transformation $X' = M\, X$, one needs to 
\begin{align}
I &= I' \notag \\
X^t\, \eta\, X &= (M\, X)^t\, \eta\, M\,X \notag \\
 &= X^t\, M^t\, \eta\, M\,X \notag \\
\Rightarrow\quad  \eta &=  M^t\, \eta\, M. \tag{*}
\end{align}
Therefore the problem is to find a set of transformations $M$ satisfying Eq. (*).
Two-dimensional case
Finding a general 4 by 4 matrix $M$ preserving the Minkowskii metric ($\eta$) requires a lot of algebra, but one can easily find the transformation preserving the 2 by 2 restriction to the  $(ct,x)$-plane.
Propose a matrix $$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$$, and solve the equation
$$
\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = 
\begin{pmatrix} a & c \\ b & d \end{pmatrix}
\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} a & b \\ c & d \end{pmatrix} =
\begin{pmatrix} -a^2+c^2 & -ab+cd \\ -ab+cd & -b^2+d^2 \end{pmatrix},
$$
which is simple if one uses the identities $\cosh^2\theta - \sinh^2\theta = 1$, and the condition $M(\theta\to 0) = \mathbf{1}$.
Thus, $$M = \begin{pmatrix} \cosh\theta & -\sinh\theta \\ -\sinh\theta & \cosh\theta \end{pmatrix}.$$
Relation with the velocity
In a similar fashion of Euclidean geometry, in which 
$$
\frac{y}{x} = \tan\theta,
$$
one uses the transformation $M$ above to relate the $ct$ coordinate with the $x$ coordinate
$$
\frac{v}{c} \equiv \frac{x}{ct} = \mathop{\mathrm{tanh}}\theta.
$$
Now,
\begin{align}
\mathop{\mathrm{tanh}^2}\theta &= 1 - \mathop{\mathrm{sech}^2}\theta \notag \\
&= 1 - \tfrac{1}{\cosh^2\theta} \notag \\
\Rightarrow\quad \cosh\theta &= \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \notag \\
\sinh\theta &= \frac{\frac{v}{c}}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}. \notag 
\end{align}
Finally, from the relation $X' = M\, X$, one obtain the usual relations
\begin{align}
x' &= -\sinh\theta\cdot t +\cosh\theta\cdot x \notag\\
 &= \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}\left( x - vt \right) \notag \\
t' &= \cosh\theta \cdot t -\sinh\theta\cdot \tfrac{x}{c} \notag\\
&= \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}\left( t - \tfrac{v}{c^2}t \right). \notag
\end{align}
