Fermat's method of calculus

The short answer is that this method works in the limit, as $\epsilon \rightarrow 0$.

Let me give you a similar example. We know from Taylor that $f(a+h) \approx f(a) + f'(a)\cdot h$

If we consider this equation for a small $h$, it makes sense. But if we simply replace $h$ by $0$, it tells us nothing.

Likewise, there is an approximation step when we replace $T(x+\epsilon)$ by $f(x+\epsilon)$ during the similarity of triangles argument.

But what is crucial is that this approximation not only works for a small $\epsilon$, but it also works in the limit, resulting in the appropriate $x$-axis cross value as $\epsilon$ gets arbitrarily close to $0$.

Edit:

We can follow the explanation in the text (page 12) as well.  From the equation $\frac{f(x_0)}{x_0-v} \approx \frac{f(x_0+e)}{x_0+e-v}$ we can solve for $v$:

$v \approx x_0 - \frac{ef(x_0)}{f(x_0+e)-f(x_0)} = x_0 - \frac{f(x_0)}{\frac{f(x_0+e)-f(x_0)}{e}}$.

If we take the limit, we get the result $v  = x_0 - \frac{f(x_0)}{f'(x_0)}$.

Now, let's think about how we would normally compute such limit for a function like $f(x)=x^2$:

$f'(x_0) = Lim_{e\rightarrow 0}\frac{f(x_0+e)-f(x_0)}{e} = Lim_{e\rightarrow 0} \frac{2ex_0 + e^2}{e} = Lim_{e\rightarrow 0} 2x_0 + e = 2x_0 $.

Notice that to compute this limit, one step was to cancel $e$ on both numerator and denominator and then replace by $e=0$.
This is what is happening in the example you mention. When solving for $v$ one $e$ literally gets canceled as a ratio, and then we take a limit.