Why does "$x^2 - 5x + 6 = 0$", which is the same as "$(x-3)(x-2) = 0$", represent a parabola?

$x^2-5x+6=0\quad$ is not the equation of the parabola.

The equation of the parabola is $\quad x^2-5x+6=y(x)$

$(x-3)(x-2)=0\quad$ is not the equation of two straight lines.

The equations of the two straight lines is $\quad x-3=y(x)\quad\text{and}\quad x-6=y(x)\quad$ or : $$(x-2-y)(x-3-y)=0$$

Do not confuse the "equation" of a curve with the "equation" to be solved for an unknown $x$.

The meaning of the word "equation" isn't the same. In the first case, it means a relationship between two variables $y$ and $x$. In the second case, it means an equality not for any values of $x$, but only for some particular values of $x$. Then, solving for $x$ means finding those particular values.

In addition :

In the very different case $\quad x^2-4xy-y^2=0\quad$ there is $y$ in the equation. This is a relationship between $y$ and $x$. So, it is valid for various values of $x$ and the related values of $y$. This allows to draw a curve $$y(x)=(2\pm \sqrt{5})\:x$$ So, two straight lines : $\quad y(x)=(2+ \sqrt{5})\:x \quad$ and $\quad y(x)=(2- \sqrt{5})\:x$

Further addition :

$x^2-5x+6=0\quad$ is commonly understood as to be solve for $x$, that is, to find the roots of the equation. The answer is two constant values : $x=2$ and $x=3$.

If one want to make understand that the question is not to find the roots of the equation on the common sens, but is to find some unknown relationship between $x$ and $y$ satisfying $x^2-5x+6=0$ , in order to avoid the ambiguity, the equation should be written as : $$\left( x(y)\right)^2-5x(y)+6=0$$ because, this specifies that $y$ exists and that the equation have to be solved for a function $x(y)$.

Solving it leads to $$x(y)-2=0 \quad\to\quad x(y)=2$$ $$x(y)-3=0 \quad\to\quad x(y)=3$$ that is two lines parallel to the y-axis.

This is because Wolframalpha is plotting $y=(x-2)(x-3)$, which is a parabola.

As you have entered $(x-2)(x-3)=0$, it is merely indicating where the intersection is between $y=(x-2)(x-3)$ and $y=0$, which is why there are red dots on the points where the $x$-coordinates are $2$ and $3$.

This is not interpreted as a function by Wolframalpha as it contains only one variable, and thus it is commonly interpreted as mentioned above, but given that $x $ is some function of $y $, it will be, as you have suggested, two lines parallel to the $y $-axis.

After much discussion in comments, I have decided to interpret this as a WolframAlpha question. Many people would not plot an equation in one variable. The solution could be plotted on a number line. In $2$ dimensions, the plot would in fact be $2$ lines parallel to the $y$-axis.

WolframAlpha does not interpret it in this fashion. It seems to interpret each side of the equation as a function in $1$ variable. $f(x)=x^2-5x+6,g(x)=0$. It graphs both and highlights points of intersection.

I believe the other answers are correct in the interpretation of the question and in the way that they have addressed the issue. But I would like to look at the question a bit differently. Whether this is of any use or interest is up to the OP.

$y = (x -2)$ and $y = (x-3)$ are lines. So why is their product a parabola?

Consider $y = (x-2)(x -3) = (x-2)x - 3(x-2)$. If you compare this to the slope-intercept form of a line, $y = mx + b$, then you get $$m = (x-2)\\b=-3(x-2)$$

In particular, it is the $m = (x-2)$ part that makes this a parabola instead of a line. In a line, the slope is constant. But in this parabola, the slope is changing in a linear fashion itself. This is not just true of this parabola, but of any parabola. That is, a parabola is just a "line" whose slope changes linearly as you go along - a relation that is more explicit when examined using calculus.

So the parabola is indeed a combination of two lines. Only the combination is by one of the lines modifying the slope of the other "line".

I get $(x-3)(x-2) = 0$. which means it represent a pair of striaght lines namely $x-2 =0 $ and $x- 3 = 0$

When we just write $$ (x-3)(x-2) = 0 $$ and ask for $x$ we usually mean the set of solutions $$ S = \{ x \mid (x-3)(x-2) = 0 \} $$

Two vertical lines in two dimensions we can model as: $$ L_1 = \{ (x,y) \mid (x-3) = 0 \} \\ L_2 = \{ (x,y) \mid (x-2) = 0 \} $$ These are different sets. We sloppily write $x-3=0$ and $x-2=0$ but mean the above.