I get a (to me) unexpected result, when doing the following computation. If anyone could point out a flaw, or maybe give some intuition, I would be grateful.
Let $a(t)$ be a smooth curve in a Lie group G. Then for an arbitrary smooth function f on G;
$$(T_{a(t)}L_{a^{-1}(t)}\dot{a}(t))f = \dot{a}(t)(f \circ L_{a^{-1}(t)}) = \left. \frac{d}{ds}\right|_{t} (f \circ L_{a^{-1}(s)} \circ a(s)) = \left. \frac{d}{ds}\right|_{t} f(e) = 0.$$
I suspect that a problem might be that $X(g) : g \mapsto T_{a(t)}L_{g}\dot{a}(t))$ is a time-dependent vector field. Thus, maybe the first equality above is invalid? How does one then interpret $(T_{a(t)}L_{a^{-1}(t)}\dot{a}(t))f$?
The flaw is in the second equality; $$ \dot{a}(t)(f\circ L_{a(t)^{-1}}) = \frac{d}{dx}\Big\vert_{s=t}(f\circ L_{a(t)^{-1}}\circ a(s)) = \frac{d}{ds}\Big\vert_{s=t}f(a(t)^{-1}a(s)) \ne 0 \quad\textrm{in general}. $$ The curve $s\mapsto a(t)^{-1}a(s)$ is a curve passing through the identity element $e\in G$ at $s=t$, and its derivative at $s=t$ is some (typically nonzero) element of $T_eG$. This is in contrast to the curve $s\mapsto a(s)^{-1}a(s)$ (as used in your calculation), which is just the identity curve $s\mapsto e$, and so has derivative $0\in T_eG$.