[proofplan]
We first spell out the translated logarithm charts and compute their inverses from the defining identity between $\exp|_U$ and $\log$. Then we identify exactly which $X\in U$ lie in the overlap of the two translated chart domains. On that open overlap, the transition map is the composition of the smooth exponential map, left multiplication by the fixed group element $g_2^{-1}g_1$, and the smooth local logarithm.
[/proofplan]
[step:Compute the inverse of each translated logarithm chart]
Fix $g\in G$. Since $V\subset G$ and $G$ is a group, the left translate
\begin{align*}
gV:=\{gv:v\in V\}
\end{align*}
is contained in $G$. The map
\begin{align*}
\psi_g:gV&\to U
\end{align*}
\begin{align*}
h&\mapsto \log(g^{-1}h)
\end{align*}
is well-defined because $h\in gV$ means $h=gv$ for some $v\in V$, hence $g^{-1}h=v\in V$.
Define
\begin{align*}
\theta_g:U&\to gV
\end{align*}
\begin{align*}
X&\mapsto g\exp X.
\end{align*}
For $X\in U$, the identity $\log(\exp X)=X$ gives
\begin{align*}
\psi_g(\theta_g(X))=\log(g^{-1}g\exp X)=\log(\exp X)=X.
\end{align*}
Conversely, if $h\in gV$, then $g^{-1}h\in V$, and the identity $\exp(\log Y)=Y$ for $Y\in V$ gives
\begin{align*}
\theta_g(\psi_g(h))=g\exp(\log(g^{-1}h))=g(g^{-1}h)=h.
\end{align*}
Thus $\psi_g^{-1}=\theta_g$, so
\begin{align*}
\psi_g^{-1}(X)=g\exp X.
\end{align*}
[/step]
[step:Identify the overlap domain in logarithm coordinates]
Let $g_1,g_2\in G$. The transition map $\psi_{g_2}\circ\psi_{g_1}^{-1}$ is defined at exactly those $X\in U$ for which $\psi_{g_1}^{-1}(X)\in g_2V$. Since $\psi_{g_1}^{-1}(X)=g_1\exp X$, this condition is
\begin{align*}
g_1\exp X\in g_2V.
\end{align*}
Multiplying on the left by $g_2^{-1}$, this is equivalent to
\begin{align*}
g_2^{-1}g_1\exp X\in V.
\end{align*}
Therefore the transition domain is precisely
\begin{align*}
D_{12}=\{X\in U:g_2^{-1}g_1\exp X\in V\}.
\end{align*}
[guided]
We need to be precise about the phrase “where defined.” A point $X\in U$ is a coordinate value for the chart $\psi_{g_1}$, and the corresponding group element is obtained by the inverse chart:
\begin{align*}
\psi_{g_1}^{-1}(X)=g_1\exp X.
\end{align*}
The second chart $\psi_{g_2}$ can be applied to this group element exactly when $g_1\exp X$ lies in the domain $g_2V$ of $\psi_{g_2}$. By the definition of the translated set $g_2V$, this means that there exists $v\in V$ such that
\begin{align*}
g_1\exp X=g_2v.
\end{align*}
Multiplying this equality on the left by $g_2^{-1}$ gives
\begin{align*}
g_2^{-1}g_1\exp X=v\in V.
\end{align*}
Conversely, if $g_2^{-1}g_1\exp X\in V$, then
\begin{align*}
g_1\exp X=g_2(g_2^{-1}g_1\exp X)\in g_2V.
\end{align*}
Thus the two conditions are equivalent, and the transition domain in the $X$-coordinates of the first chart is exactly
\begin{align*}
D_{12}=\{X\in U:g_2^{-1}g_1\exp X\in V\}.
\end{align*}
[/guided]
[/step]
[step:Derive the transition formula on the overlap]
For $X\in D_{12}$, the preceding step ensures that $g_2^{-1}g_1\exp X\in V$, so the logarithm below is defined. Using the formula for $\psi_{g_1}^{-1}$, we compute
\begin{align*}
(\psi_{g_2}\circ\psi_{g_1}^{-1})(X)=\psi_{g_2}(g_1\exp X).
\end{align*}
By the definition of $\psi_{g_2}$,
\begin{align*}
\psi_{g_2}(g_1\exp X)=\log(g_2^{-1}g_1\exp X).
\end{align*}
Therefore
\begin{align*}
(\psi_{g_2}\circ\psi_{g_1}^{-1})(X)=\log(g_2^{-1}g_1\exp X)
\end{align*}
for every $X\in D_{12}$.
[/step]
[step:Prove that the transition domain is open and the transition map is smooth]
Define the map
\begin{align*}
F:U&\to G
\end{align*}
\begin{align*}
X&\mapsto g_2^{-1}g_1\exp X.
\end{align*}
The matrix exponential map is smooth on $\mathfrak g$ in these exponential coordinates by the local exponential chart hypothesis, equivalently by [citetheorem:8791] near $0$. Left multiplication by the fixed element $g_2^{-1}g_1\in G$ is smooth because the group multiplication map of the matrix Lie group is smooth. Hence $F$ is smooth, and therefore continuous.
Since $V$ is open in $G$, the preimage
\begin{align*}
F^{-1}(V)=\{X\in U:g_2^{-1}g_1\exp X\in V\}=D_{12}
\end{align*}
is open in $U$. On $D_{12}$, the transition map is
\begin{align*}
X\mapsto \log(F(X)).
\end{align*}
The map $F|_{D_{12}}:D_{12}\to V$ is smooth, and $\log:V\to U$ is smooth by hypothesis. Hence $\log\circ F|_{D_{12}}$ is smooth. This is exactly $\psi_{g_2}\circ\psi_{g_1}^{-1}$ by the transition formula, so the transition map is smooth.
[/step]
