I have the group $\langle a,b,c\rangle/\langle -b+c-a,b+c-a\rangle$. I know this is $\mathbb{Z}\oplus\mathbb{Z_2}$. However, I tried doing it like this and got something else :
I have $$-b+c-a=0, b+c-a=0$$ Which gives me $2c=2a$ and $2b=0$. Then my group is the same as $$\langle a-c,b,c\rangle/\langle 2(a-c),2b\rangle$$ which is again $$\langle d,b,c\rangle/\langle2d,2b\rangle\simeq\mathbb{Z}\oplus\mathbb{Z_2}\oplus\mathbb{Z_2}$$ Which step was wrong?
The group is not the same as $\langle a - c, b, c \rangle / \langle 2 (a-c),2b \rangle$. This is because you did not apply invertible transformations to your relations. The relations $c = a + b$ and $c = a-b$ are equivalent to your given relations, but they do not follow from $2c = 2a$ and $2b = 0$.
What we do get is the abelian group presentation $$\langle a, b, c \rangle / \langle a + b -c, a-b-c \rangle$$
Substituting $a' = a+b$ (so $a-b = a' -2b$) yields that your group is isomorphic to $$\langle a', b, c \rangle / \langle a' -c, a'-2b-c \rangle \cong \langle b, c \rangle / \langle 2b \rangle \cong \mathbb Z_2 \oplus \mathbb Z.$$
In general, to tell whether your substitutions are invertible, you can identify the $n$ generators of an abelian free group with basis vectors of $\mathbb Z^n$, and check whether your new generators are still a basis of $\mathbb Z^n$. This can by done by evaluating the determinant of the matrix whose columns are the new candidate generators.
In this example, we have three generators $a,b,c$, identified with the vectors $(1,0,0), (0,1,0), (0,0,1) \in \mathbb Z^3$. My substitution $a' = a + b$ corresponds to the change of basis matrix $$\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},$$ which has determinant $1 \in \mathbb Z^*$, so is invertible over $\mathbb Z$.
Similarly, if you want to check whether two sets of relations are equivalent, you can, sloppily speaking, check whether they generate the same vector space. In your example, the relations $2c -2a, 2b$ correspond to the vector space spanned by $(-2, 0 2), (0,2,0)$, which is not the same as the vector space spanned by $(1,1,-1),(1,-1,-1)$ corresponding to $a+b-c,a-b-c$ (one way to check this is to note that in the first case, all vectors have even entries and in the second case this is not true).