Here's a useful general result.
Theorem.
Suppose $X$ is a random variable with cumulative distribution function $F_X$, let $a, b \in \mathbb{R}$ with $b > 0$, and let $Y = a + b X$.
- The cumulative distribution function of $Y$ (call it $F_Y$) is given by
$$
F_Y(y) = F_X\left(\frac{y - a}{b}\right)
$$
for all $y \in \mathbb{R}$.
- If $X$ is also continuous with density $f_X$, then the density of $Y$ (call it $f_Y$) is given by
$$
f_Y(y) = \frac{1}{b} f_X\left(\frac{y - a}{b}\right)
$$
for all $y \in \mathbb{R}$.
Your question is the special case where $X \sim \chi^2_1$ (whose density you know), and you're interested in the density of $Y = \frac{1}{2} X$ (here $a = 0$ and $b = 1/2$).
The theorem above will allow you to compute the density of $Y$.
Proof of 1.
Just compute:
$$
\begin{aligned}
F_Y(y)
&= P(Y \leq y) \\
&= P(a + b X \leq y) \\
&= P\left(X \leq \frac{y - a}{b}\right) && \text{(*)}\\
&= F_X\left(\frac{y - a}{b}\right).
\end{aligned}
$$
(*) Here we used the assumption that $b > 0$. If $b < 0$, the inequality would change direction, and if $b = 0$, we wouldn't be able to divide by $b$ at all.
Proof of 2.
Differentiating the result of the first part, we get
$$
\begin{aligned}
f_Y(y)
&= \frac{d}{dy} F_Y(y) &&\text{(**)}\\
&= \frac{d}{dy} F_X\left(\frac{y - a}{b}\right) \\
&= \frac{1}{b} F^\prime_X\left(\frac{y - a}{b}\right) &&\text{(chain rule)}\\
&= \frac{1}{b} f_X\left(\frac{y - a}{b}\right). &&\text{(**)}
\end{aligned}
$$
(**) Here we used the fact that a density of a continuous random variable is the derivative of that random variable's cumulative distribution function, which is a consequence of the definition of the density and the fundamental theorem of calculus.