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.
 A: Consider two observers $O$ and $O'$ moving with velocity $v$ with respect to each other. Both of them will use a photon bouncing off a mirror to define any kind of duration, so it is easy to show that their duration will satisfy the relation
$$\Delta t' = \gamma \Delta t \,,$$ 
and a similar argument with measuring any relative distance $\Delta l$ between points in the direction of relative movement gives
$$\Delta l' = \gamma \Delta l \, .$$
The stated relations will hold under any coordinate convention. Now suppose observer $O$ decides to establish a coordinate convention by setting $t=t'=0$ when her coordinate origin $x=0$ met with the coordinate origin $x'=0$ of $O'$. She then asserts that the relative distance between their origins grows as $-vt$, so they have after transforming any relative distances
$$x'= \gamma (x - vt) \, .$$
By the same argument, she concludes that $O'$ must see the relative distances as changing in the opposite way with $+v t'$ and after using the same transform (due to the relativity principle), $O$ concludes that
$$x = \gamma (x' + v t'\!) \,.$$
But, as we know, the relative growth $\Delta t$ is not the same as $\Delta t'$. By substituting for $x'$ and turning the last equation into an expression for $t'$ we have. 
$$t'= \gamma(t - \frac{vx}{c^2})$$
I.e., it is a question of consistence of the coordinate convention picked out by $O$ that we need this correction $vx/c^2$. If we chose $O'$ to fix her origin on the one of $O$, we would get
$$x'=\gamma x,\, t'=\gamma t \,.$$
The term in the transformation is thus a consequence of the fact that we choose to mix a $vt$ factor into the coordinate convention which in turn requires a skewing of time through space. It is very important during the discussion of special relativity to remember that this is a conventional (non-physical) coordinate transform fix.

EDIT: As requested, I will present a direct derivation which gives first a conventional meaning to simultaneity (the $t=\rm const.$) before requesting the consistency of the $x$ coordinate and without any explicit statement of length contraction. We only have to presume we already know the standard derivation of the relation
$$\Delta t' = \frac{\Delta t}{\sqrt{1- v^2/c^2}}$$
Consider the following simultaneity convention: 
Consider an array of clocks in rest with respect to $O$, this is a fixed observable feature. A light signal is sent from the origin of $O$ (this is the conventional point, the referential signal point could be anywhere in the coordinate system, even moving with respect to it) denoting the $t=0$ moment. 
All the clocks set their time to $t=0$ upon receiving the signal even though there is a certain delay $\delta t_C$ for every individual clock $C$ and immediately send a light signal with their identification back to $O$. $O$ receives the signal from a clock $C$ at a time $t_C$.
She naturally concludes that $t_C =  \Delta l/c  +  \Delta l/c$ and from that computes that when she sends a signal to $C$, the signal will be there at a delay $\delta t_C =t_C/2$. She then sends a signal to $C$ at her own time $t$ to set itself to $t + \delta t_C$. Once this process is finished for every clock $C$, a notion of the time coordinate $t$ or simultaneity is established globally for all points seen by $O$.
But consider now that $O$ watches $O'$ while establishing a similar convention with respect to a similar set of clocks, i.e., a set of clocks which is moving with a velocity $v$ in the $x$ direction with respect to $O$ but is in rest with respect to $O'$. $O'$ sends out the signal at $t'=0$, but the clocks in front of her (in the direction of motion) are "rushing away from the signal" with the speed of light constant, so the delay will be longer. On the contrary, the clocks behind $O'$ are "rushing towards the signal", so the delay is shorter. For clocks strictly in front of and behind $O'$, we have a delay
$$\delta t_{b/f} = \frac{\Delta l}{c \pm v}$$ 
Once the clocks send their signal back, an opposite effect takes place for the clocks in front/ behind. The resulting time the second clocks send back their signals is thus identical for both cases (as measured by the clock system established by $O$):
$$t_C = \frac{\Delta l}{c + v} + \frac{\Delta l}{c - v} = \frac{2 \Delta l}{c}\frac{1}{1 - v^2/c^2}$$
$O$ now uses the knowledge that $O'$ actually measured this time delay with a dilation, so she knows $O'$ will be sending out a correction time 
$$\delta t'_C  = \frac{ \Delta l}{c}\frac{1}{\sqrt{1 - v^2/c^2}}$$
$O$ is perfectly clear on the fact that the duration from the point of view of $O'$ $t'$ since the time they met will be rescaled by the factor $1/\sqrt{1-v^2/c^2}$.  However, $O$ concludes that also the sent out time/correction should be different to conform to her view of simultaneity. $O$ sees the duration of flight of the signal to be $\Delta l/(c \pm v)$ from her own point of view, but also realizes the clocks co-moving with $O'$ run slower, so she has to multiply the correction by $\sqrt{1-v^2/c^2}$ to compensate for the fact. She then sees a total discrepancy between her simultaneity and the $O'$ simultaneity of the magnitude
$$\delta t'_{b,f} - \delta t'_C = \frac{\Delta l}{c}\left(\frac{\sqrt{1- v^2/c^2}}{1 \pm v/c} - \frac{1}{\sqrt{1-v^2/c^2}} \right) $$
Which gives after a bit of algebra
$$\delta t'_{b,f} - \delta t'_C = \mp \frac{v \Delta l}{c^2} \frac{1}{\sqrt{1- v^2/c^2}}$$
Using a coordinate $x$ which is 0 at $O$, and the positive sign means "in front" and negative "behind", we can conclude that $O$ will summarize the "deformation" of $O'$ global notion of time as
$$t' = \frac{t}{\sqrt{1-v^2/c^2}} - \frac{ v x}{c^2} \frac{1}{\sqrt{1- v^2/c^2}} = \gamma (t - vx/c^2)$$
$\blacksquare$
A: 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}
