Deriving the Lorentz Transformation

I'll not derive the transformation (that has been done in countless books and articles, I am sure you can find them yourself) but instead will try to explain why the formula you propose can't be correct.

For starters, observe that since you don't touch $y$ and $z$, we might as well work in 1+1 dimensions. Also, let $c=1$ so that we aren't bothered by unimportant constants (you can restore it in the end by requiring that formulas have the right units). Then it's useful to reparametrize the transformation in the following way $$x' = \gamma(x - vt) = \cosh \eta x - \sinh \eta t$$ $$t' = \gamma(t - vx) = -\sinh \eta x + \cosh \eta t$$ where we introduced rapidity $\eta$ by $\tanh \eta = v$ and this by standard (hyperbolic) trigonometric identities implies $\cosh \eta = \gamma = {1 \over \sqrt{1 - v^2}}$ and $v \gamma = \sinh \eta$, so that this reparametrization is indeed correct.

Now, hopefully this reminds you a little of something. In two-dimensional Euclidean plane we have that rotations around the origin have the form $$x' = \cos \phi x + \sin \phi y$$ $$y' = -\sin \phi y + \cos \phi x$$ and this is indeed no coincidence. Rotations preserve a length of vector in Euclidean plane $x'^2 + y'^2 = x^2 + y^2$ and similarly, Lorentz transformations preserve space-time interval (which is a notian of length in Minkowski space-time) $x'^2 - t'^2 = x^2 - t^2.$ You can check for yourself that only the stated transformation with hyperbolic sines and cosines can preserve it and consequently the change you introduced will spoil this important property. Also, if you are familiar with phenomena like relativity of simultaineity, one could also argue on physical grounds that your proposed change can't lead to physical results.

Incidently, there has recently been asked similar question to yours, namely how to derive that the transformation is linear purely because of the preservation of space-time interval. You might want to check it out too.

You should look at this answer, because it derives the term you want right away. Einstein's postulates $\leftrightarrow$ Minkowski space for a Layman

The reason it's $t'=(t-vx)/\sqrt{1-v^2}$ and not $t'=t/\sqrt{1-v^2}$ (you must set c=1 to follow anything in relativity) is simple--- it's failure of simultaneity at a distance. The coordinate lines t=constant can't stay horizontal in a space-time diagram--- they have to get tilted up by the same amount that the time axis is tilted right. The remaining factors can be understood by reproducing time-dilation and length-contraction arguments, but failure of simultaneity is the most important nonintuitive effect, and it is the first discussed by Einstein in his paper, for this reason.

The form of the Lorentz transformation should be constrasted with the form of a rotation of the x and y coordinates, so that the x coordinate gets a slope of m:

$$x' = { x+my \over\sqrt{1+m^2}}$$ $$y' = { y-mx \over\sqrt{1+m^2}}$$

or if you use different units for x and y, say x in inches and y in centimeters,

$$x' = { x + my \over \sqrt{1+{m^2\over c^2}} }$$ $$y' = {y - {mx\over c^2}\over \sqrt{1+{m^2\over c^2}} }$$

Where c is a universal constant of nature: the isoceles slope of right, which is the slope of an isoceles right triangle with legs along the x and y axis. It's magnitude is 2.54 cm/inch.

I also have struggled with this issue for sometime. So, here's the answer completely intuitive. It is obviously related with relativity of simultaneity. Now in the moving frame, if two events are simultaneous for the moving person, don't make it simultaneous for you. If you work out the calculations, you'll see that there is a time difference of $(vx/c^2)\gamma$, where the event farther in the direction of motion happens later by this much time difference in the rest frame. So you add this factor to the time of the clock of the observer in case of the fact that the person is moving along the positive side of your origin of spacetime

Now, why the negative sign in the reverse transformation, well it turns out that this is an interplay of choosing a right handed coordinate system or saying that all observers, even those moving in the opposite direction, must attribute the same side of origin as positive in their coordinate systems. This was implicitly done even in gallilean days when you solved simple kinematics problems.

Now what happens is in the negative velocity case, if you assign coordinates to events, the one which is more farther along your direction of motion now will be negative too. Hence, the 'x' coordinate will be negative, and velocity is also negative and two negatives make a positive. So, this again leads to addition of extra time for the event, since the clock tick happened earlier than the event you were recording.

All this can be rederived if you think carefully about two simulatenous events in a moving frame. I hope this clears it up.