I have a really basic question which I am struggling to articulate formally in differential geometry. I have a Lie group $G$, and a tangent vector $v\in T_{1_G}(G)$. I want to claim that the vector field $g\mapsto(g,v)$ is a smooth vector field, but I am unsure as to whether I can say formally that $v\in T_g(G)$ for all $g\in G$.
With $G$ being a topological group, it is homogeneous, and given a chart $(U,\varphi)$ of the identity, $(gU, g^{-1} \cdot \varphi)$ is a chart around any $g$. But I'm still unsure as whether that means that $v\in T_g(G)$.
If it is trivially true that $v\in T_g(G)$ for $g$ in the same component as $1_G$, or if it true even without the topological group sturcture, I would be very thankful if someone were to help my overcome this conceptual hurdle.
The concept you're looking for here is left-invariant vector field. For Lie group $G$ and any $v \in T_1 G$, we can define an (a priori, rough) vector field $$\tilde v \in \Gamma(TG), \qquad \tilde v_g := T_1 L_g \cdot v .$$ Here, $L_g : G \to G$ is the left multiplication map $h \mapsto gh$. Then, unwinding definitions shows that $\tilde v$ is invariant under left multiplication, that is that $$T_1 L_g \cdot \tilde v_h = \tilde v_{gh}$$ for all $g, h \in G$.
Conversely, there is canonical map $$\omega : TG \to T_1 G , \qquad \omega : w_g \mapsto T L_{g^{-1}} \cdot w_g ,$$ and unwinding definitions gives that $$\omega(\tilde v_g) = v.$$ If we identify $T_1 G$ with the Lie algebra $\mathfrak g$ of $G$, $\omega$ is precisely the Maurer-Cartan form on $G$. This in turn defines a canonical bundle isomorphism $$\Phi : TG \to G \times T_1 G, \qquad \Phi : w_g \mapsto (g, \omega(w_g)) .$$ This bundle isomorphism formalizes the notion in the original question: Unwinding definitions shows that $\Phi^{-1}$ identifies $(g, v) \in G \times T_1 G$ with $\tilde v_g \in T_g G$.
With this in hand, we can check smoothness directly by verifying that $\tilde v \cdot f$ is smooth whenever $f$ is, and this again amounts to unwinding definitions and recognizing that an expression for $\tilde v \cdot f$ is manifestly smooth.