Studying distribution and density of absolute value of a r.v Let $X$ have dsitribution $F$. How can we find distribution of $Y = |X| $.
Suppose $X$ admits a density $f_X$, then $Y$ also admits a density $f_Y$:
I was able to do this and I got $$f_Y(y) = f_X(y) 1_{(0, \infty)} + f_X(-y) 1_{ (- \infty, 0 ] } $$
is this correct? and how can I find the distribution?

Close:
You are using the transformation formula: $f_Y(y) = f_X(x(y))\left\lvert {\frac{\mathrm d x(y) }{\mathrm d y\quad}}\right\rvert$
And you have correctly identified that as $y(x)=\lvert x\rvert$ does not strictly have an inverse function, there are two $x$ functions for each $y$ value.  $x_1(y)=y$ and $x_2(y)=-y$ both for all $y$ in $[0; \infty)$.
So you should have:
$$f_Y(y) = f_X(y) \;\mathbf 1_{[0;\infty)}(y) + f_X(-y)\;\mathbf 1_{(0;\infty)}(y)\quad$$
Where the indicator function is: $\mathbf 1_A(y) = \begin{cases} 1 & : y\in A\\0 & : y\notin A\end{cases}$
Then by integration you can show: $$F_Y(y) = \Big(F_X(y) - F_X(-y) \Big)\;\mathbf 1_{[0;\infty)}(y)$$
Which makes sense since: $$\mathsf P(0\leq Y\leq y) =\mathsf P(-y\leq X\leq y)$$
