How do we find the value of dimension of a vector space?
My teacher said it's the number of free variables in the echelon form of the matrix. But according to some articles, it is the number of columns with pivot element in it (i.e: the number of elements in the basis).
For example, my teacher said that dimension of the vector space under the following conditions is 1, but I think it should be 2.
$$ \begin{align*} x + y + z &= 0 \\ x + 2y &= 0 \\ y - z &= 0 \end{align*} $$
What you should do with "the matrix" depends on how exactly your matrix relates to the space in question. In particular, your space is described by the equations $$ x + y + z = 0\\ x + 2y = 0\\ y - z = 0 $$ which is to say: your space is the set of all solutions to $Ax = 0$, where $$ A = \pmatrix{1&1&1\\1&2&0\\0&1&-1} $$ That is, your space is the null space of the matrix $A$. The dimension of a matrix's null space is the number of free variables in its echelon form.
On the other hand, we might want to describe a space differently. For example, we might define the a new space that consists of all vectors of the form $$ (x+y+z,x+2y,y-z) $$ which is to say that our space is the set of all vectors of the form $Ax$ (for the same $A$). In other words, our space is the column space of $A$. The dimension of a matrix's column space is the number of columns in the echelon form that have a pivot element.
In the end, your teacher is right: the dimension of your vector space is $1$, not $2$.
To supplement @Omnomnomnom's answer, I have worked out calculating the dimension of the vector space below.
The dimension of a vector space is defined as the number of elements (i.e: vectors) in any basis (the smallest set of all vectors whose linear combinations cover the entire vector space).
In the example you gave, $x = -2y$, $y = z$, and $z = -x - y$. So,
$$ \begin{align*} \begin{pmatrix}x\\y\\z\end{pmatrix} &= \begin{pmatrix}-2y\\z\\-x - y\end{pmatrix} \\&= \begin{pmatrix}-2z\\z\\-x - z\end{pmatrix} \\&= \begin{pmatrix}-2z\\z\\z\end{pmatrix} \\&= z\begin{pmatrix}-2\\1\\1\end{pmatrix}. \end{align*} $$
Thus, we can say the vector space under the conditions you gave is given by $\left\{ x \begin{pmatrix}-2\\1\\1\end{pmatrix} : x \in \mathbb{R} \right\}$ (i.e: the linear combinations of the vector $(-2, 1, 1)$; in this case, since there's only one vector, linear combinations are simply multiples of that vector).
So, the basis of this vector space is $\left\{ \begin{pmatrix}-2\\1\\1\end{pmatrix} \right\}$. There is only one vector in the basis, so the dimension of the vector space is $1$.
To supplement the already given answers, let me add some general considerations. When you ask
How do we find the value of dimension of a vector space?
you have to be much more specific — which vector space are we talking about? It's like going to a grocery store and literally asking a clerk "How much does it cost?" Most likely the clerk will respond "How much what costs?" They have lots of products at different prices, and to get an answer you have to be specific as to what your inquiry is about.
So generally speaking, nobody can give an answer to your question (as quoted above) until you tell us which vector space we're talking about. Even when given a matrix, there are several vector spaces associated with it, most notably (but not only) the column space and the null space. They are two different things with different dimensions and different methods for finding them. That's what your confusion was — you were talking about two different things as if they were the same, thus confusing them with each other.
As for the specific example (of a system of equations) in the second part of your question, it's already been addressed by others' great responses.