If we take the definition that a covering map $\Pi \colon \tilde{G}\to G$ to be a homomorphism of connected Lie groups such that the induced map on Lie algebras $\pi\colon \mathrm{Lie}(\tilde{G})\to \mathrm{Lie}(G)$ is an isomorphism, then the fact that $\Pi$ induces an isomorphism between $\tilde{G}/\ker(\Pi)$ and $G$ can be seen as follows:
First, the induced map $\Pi \colon \mathrm{Lie}(\tilde{G})\to \mathrm{Lie}(G)$ can be viewed as the map induced by $\pi$ on the tangent spaces of the identities $e_{\tilde{G}}$ and $e_G$. If this map is an isomorphism, then the inverse function theorem implies that $\pi$ is locally a diffeomorphism near $e_{\tilde{G}}$, that is, there is an open neighbourhood $N$ of $e_{\tilde{G}}$ such that $\pi_{|N}\colon N \to \pi(N)\subseteq G$ is a diffeomorphism. In particular $\ker(\Pi)\cap N = \{e_{\tilde{G}}\}$, so that $\ker(\Pi)$ is a discrete subgroup of $\tilde{G}$, and hence if we let $G_1 = \tilde{G}/\ker(\Pi)$, the quotient map $Q\colon \tilde{G} \to G_1$ induces an isomorphism $q\colon \mathrm{Lie}(\tilde{G}) \to \mathrm{Lie}(G_1)$. Since $\Pi$ induces a map $\Pi_1 \colon G_1 \to G$ it follows we are reduced to showing the following:
Claim: If $\Phi\colon G_1 \to G$ is a homomorphism of connected Lie groups which has trivial kernel and which induces an isomorphism $\phi\colon \mathrm{Lie}(G_1)\to \mathrm{Lie}(G)$ then $\Phi$ is an isomorphism.
Proof of claim: By the discussion above, there is an open neighbourhood $N_1 \subseteq G_1$ of $e_{G_1}$ such that $\Phi_{|N_1}\colon N_1\to N$ is a diffeomorphism, where $N$ is an open neighbourhood of $e_G\in G$. If $g\in \tilde{G}_1$ is an arbitrary element of $G_1$ and we let $\lambda_g\colon \tilde{G}\to \tilde{G}$ be the map $\lambda_g(h_1) = g.h_1$, $(\forall h_1 \in G_1$) then $\lambda_g$ is a diffeomorphism with inverse $\lambda_g^{-1} = \lambda_{g^{-1}}$. Similarly if $k = \Phi(g)\in G$, then $\lambda_{k}\colon G \to G$ given by $\lambda_{k}(h) = kh$, ($\forall h \in G$) is a diffeomorphism from $G$ to itself, hence
if we set $N_{1,g} = \lambda_g(N_1) =g.N$, then if $x \in N_{1,g}$ we have
$$
\Phi(x) = \Phi(g.(g^{-1}.x))=\Phi(g).\Phi(g^{-1}.x)= \lambda_{k}\circ \Phi_{|N_1} \circ \lambda_g^{-1}(x)
$$
so that $\pi_{|N_g}$ is a composition of diffeomorphisms, and hence a diffeomorphism from $N_{1,g}$ to $N_k = k.N$. But then $\Phi(G_1) = \bigcup_{g \in G_1} \Phi(g.N_1) = \bigcup_{g\in G_1} \Phi(g).N$ is an open subset of $G$. But then if we set $H=\Phi(G_1)$, $\Phi\colon G_1 \to H$ is diffeomorphism from $G_1$ to $H$, where $H$ is an open subgroup of $G$. But $G=\bigsqcup_{g \in G/H} g.H$ is the disjoint union of its left $H$-cosets, which implies that each coset is both open and closed, and thus as $G$ is connected, we must have $H=G$ as required.
