First of all, parameters of unweighted linear regression are very strongly influenced by the largest values of $y$, the dependent variable. When you linearize an exponential model and you have very small values, then the $\log(y)$ can be very large (negative) and strongly influence the results.
I made an example fo your problem.
I generated $20$ equally spaced $x$'s $(x_i=1,2,3,\cdots,20)$ and generated the corresponding $y$ using $$y=e^{5.678-0.456 x}$$ to which I added a random noise for a maximum relative error of $\pm10$%.
For the first run, I just linearized the system and used the classical linear regression. The resulting parameters are $5.710$ and $-0.460$. For these values was computed $$SSQ=\sum_{i=1}^{20} \Big(y_i^{calc}-y_i^{exp}\Big)^2$$ which is equal to $87.07$.
In a second step, I used the weighting as suggested in the link you give. The resulting parameters are $5.655$ and $-0.443$ and $SSQ=26.64$.
In a third step, starting with the parameters obtained by the first step, I performed a nonlinear regression. The resulting parameters are $5.656$ and $-0.444$ and $SSQ=26.54$.
As you can see, what is proposed leads to results which are quite close to the nonlinear regression.
I must confess that I always perform a nonlinear regression even if the model can be linearized and even if errors are small. Remember that what is measued is $y$ and not $\log(y)$ and when you compare sum of squares they must be consistent.
By the way, you could demonstrate easily that, if the errors are not too large, minimizing the sum of the squares of logarithms is almost identical to minimizing the sum of squares of relative errors. For illustration purposes, I also made that. The resulting parameters are $5.706$ and $-0.460$ and $SSQ=74.74$. Compare these results to those from the first step.
The concepts of "leverage" point and "influenctial" point are not equivalent. High leverage in regression analysis refers to observations that are outlying values of the independent variables. More technically, high leverage points can be defined as those having no neighbouring points in a $R^n$ space, where $n$ is the number of independent regression variables. When leverage points occur in regression analyses, the fitted model commonly passes relatively close to them. As a result, high leverage obsevations have the potential to yield large changes in the parameter estimates when they are deleted. This explains why influential points "often" have high leverage, and vice versa. However, it is common to find observations where this is not true.
A high leverage point is not necessarily an influenctial point. The better way to understand this is to imagine a point at the extreme ranges of independent variables, but that is well aligned to the fitting model obtainable by all other observations: this point would clearly have high leverage, but its deletion would have a small effect on the model. In other words, it would be an outlier for the independent variables, but not an influenctial point.
On the other hand, a low leverage point is not necessarily a non-influenctial point. A way to understand this is to imagine a point at the middle ranges of independent variables, but that is not aligned to the fitting model obtainable by all other observations (e.g., an outlier for the dependent variable). This point would have low leverage, but its deletion could have a considerable impact on the regression model.
Best Answer
You shouldn't ask so many questions at once. That said: