Derivation of the pdf of RV $X (\left|X \right| + 1)$ How to derive the probability density function $p_Y(y)$ in case, when $Y = X (\left|X \right| + 1)$, where $X  \sim \mathrm{Uniform}(-1, 1)$.
I know that if $X < 0$ and $(\left|X \right| + 1) X < y$ then $X < \dfrac{1}{2} \left(1- \sqrt{1 - 4y} \right)$ and also if $X \geq 0$ and $(\left|X \right| + 1) X < y$ then $X < \dfrac{1}{2} \left(\sqrt{1 + 4y}  - 1 \right)$. But I do not know how to find CDF in this case. How correctly to combine these conditions into one?
Thank you very much!

Your results so far are correct and you are almost there. 
Note that the density of $X$ is symmetric with respect to zero as in $\Pr\{ X < 0 \} = \Pr\{ X > 0 \}  = \frac12$. Meanwhile, $X < 0$ actually means $-1 < X < 0$ by definition, and similarly $X > 0$ implicitly means $0 < X < 1$.
Also note that $Y$ has the same sign as $X$. That is, $Y < 0$ whenever $X < 0$ and when $X$ is positive then so is $Y$.
Last thing to note is that the density of $Y$ is also symmetric with respect to zero. This is due to $|X|+1$ having the range of $[1,2]$ and the product with the symmetric $X$ just produces a copy on the negative side. Thus we already have $\Pr\{ Y < 0 \} = \Pr\{ Y > 0 \}  = \frac12$ before doing any calculation.

On the negative $X$ side, we have
\begin{align}
\Pr\bigl\{ Y < y ~\big| ~X < 0 \bigr\} &= \Pr\Bigl\{ -1 < X < 0 ~~\& ~~X < \frac{1}{2} \left(1- \sqrt{1 - 4y} \right)\Bigr \} \\
& = \frac{1}{2} \left(1- \sqrt{1 - 4y} \right) - (-1)
\end{align}
where you already knew that $r_m \equiv \frac12 \left(1- \sqrt{1 - 4y} \right)$ is smaller than zero, hence we do the $r_m -(-1)$ to take the length from the left end point (negative one) to $r_m$ as the probability for the uniform distribution.
Putting $\Pr\{ X < 0 \}$ back, unconditionally we have on the negative $Y$ part (since $Y$ follows the sign of $X$)
$$F_Y(y) = \frac12 \cdot \frac12 \left( 3 - \sqrt{1 - 4y} \right) \qquad \text{for}~~ y < 0$$
On the positive side, the conditional CDF (ignoring the negative side) is
\begin{align}
\Pr\bigl\{ Y < y ~\big| ~X > 0 \bigr\} &= \Pr\Bigl\{ 0 < X < 1 ~~\& ~~X < \frac12 \left(-1 + \sqrt{1 + 4y} \right) \Bigr \} \\
& = \frac12 \left(-1 + \sqrt{1 + 4y} \right) - 0
\end{align}
where we take the length between the left end point (now zero) and $r_p \equiv \frac12 \left(-1 + \sqrt{1 + 4y} \right)$ which you already knew is between $0$ and $1$.
Now, unconditionally we are no longer ignoring the negative side and have to put things on top of it. Namely, we already have accumulated the $1/2$ probability mass from $y<0$.
\begin{align} 
F_Y(y) &= \Pr\bigl\{ Y < \frac12 \bigr\} + \Pr\bigl\{ X > 0 \bigr\} \cdot \Pr\bigl\{ Y < y ~\big| ~X > 0 \bigr\} \\
&= \frac12 + \frac12 \cdot \frac12 \left(-1 + \sqrt{1 + 4y} \right) \\
& = \frac14 \left( 1 + \sqrt{1 + 4y} \right)  \qquad \text{for}~~ y > 0
\end{align}

As mentioned before, the density of $Y$ is symmetric (even) while the cumulative function is rotationally symmetric (odd) with respect to $(0,\frac12)$. This can be emphasized by using the sign function $Sgn(\cdot)$, which is odd, and the absolute value that is even.
\begin{align}
f_Y(y) &\equiv \frac{d F_Y(y) }{ dy } = \frac1{2 \sqrt{ 1 + 4 |y| } } & &\text{for} ~~ -2 < y < 2  \\
F_Y(y) &= \frac12 - \frac{ Sgn(y) }4 \Bigl( 1 - \sqrt{1 + 4 |y|} \Bigr)  & &\text{for} ~~ -2 < y < 2
\end{align}
The density written in this form is obviously an even function. As for the CDF, note that the sign fucntion (odd) multiplying the big parenthesis (even) is odd.
