It's easier to understand the injectivity when $X=\tilde{X}/G$ where $G$ is a finite group acting freely on $\tilde{X}$. The image of $\pi^*: \Omega^k(X)\to\Omega^k(\tilde{X})$ is the space $\Omega_G^k(\tilde{X})$ consisting of $G$-invariant forms, i.e., forms $\omega\in\Omega^k(\tilde{X})$ such that

$$ g^*\omega=\omega,\;\;\forall g\in G.$$

Note also that the resulting map

$$\pi^*: \Omega^k(X)\to\Omega^k_G(\tilde{X})$$

is a bijection. Suppose $\omega\in\Omega^k(X)$ is a closed form on $X$ such that $\pi^*\omega$ is exact, i.e.,

$$\pi^*\omega= d\tilde{\eta}.$$

Set

$$\bar{\eta}:=\frac{1}{|G|}\sum_{g\in G} g^*\tilde{\eta}$$

By construction $\bar{\eta}$ is $G$-invariant and

$$ d\bar{\eta}= \frac{1}{|G|} \sum_{g\in G} g^*d\tilde{\eta}= \frac{1}{|G|}\sum_{g\in G} g^* \pi^* \omega =\pi^* \omega $$

since $\pi^*\omega $ is $G$-invariant.

Since $\bar{\eta}$ is $G$-invariant, there exists $\eta\in \Omega^{k-1}(X)$ such that $\pi^*\eta=\bar{\eta}$. Hence

$$ \pi^* \omega=d\pi^* \eta \Rightarrow \pi^*(\omega-d\eta)=0.$$

Since $\pi^*:\Omega^k(X)\to\Omega^k(\tilde{X})$ is injective we deduce $\omega=d\eta$ , i.e., a closed form on $X$ is exact if and only if its pullback to $\tilde{X}$ is exact.