Definition of Left Invariant Vector Fields The definition (according to Marsden and Ratiu's $\textit{Introduction to Mechanics and Symmetry}$) of a left invariant vector field $X$ on a Lie group $G$ states that for each $g \in G$, $L_g^*X=X$. That is, $$(T_hL_g)X(h)=X(gh)$$ for every $h\in G$. Other sources put $(dL_g)_h$ in place of $T_h$ (Jack Lee's $\textit{Intro to Smooth Manifolds}$ follows this latter convention). My question is regarding the notation $T_h$ and the result $X(gh)$. Why do we write $T_h$ as the partial derivative w.r.t $h$ and what does it actually mean to take the derivative of the left action as we are doing in this definition? It seems we start with a vector at $h$ and map to a vector at $gh$ but these books just state it and move on so some clarification would be helpful.  
 A: Let $g, h \in G$. 
Both $T_h$ and $(d \  \cdot)_h$ denote taking the tangent map at the point $h \in G$. Notably we do have any vector space structure on the Lie-Group $G$. Therefore naively defining derivatives via differential quotients is not possible, because differences such as $g - h$ are not defined without a vector space being involved. 
The tangent map $T_h L_g$ at $h$ of $L_g \colon \ G \to G$ maps between the tangent spaces $T_h G$ at $h$ and $T_{L_g h} G = T_{gh}G$ at $gh$. 
Both concepts are covered in earlier chapters of Lee's book. That's why he just moves on unceremoniously. So you should consult those if you want to understand things rigorously.  
A: It is not really a partial derivative, it is first of all the differential or tangent map of a smooth map on a manifold, which is evaluated at a point. In this case, it is $L_g$, and its differential gets evaluated at a point $h$, resulting in a linear map between the corresponding tangent spaces $T_h G$ and $T_{L_g(h)}G = T_{gh}G$. Wether you write $(d L_g)_h$ or $T_h L_g$ is just a matter of notation. Equivalently $X$ is left invariant iff
$$ (dL_g)_h X_h = X_{gh}$$ for any $g,h \in G$, where I denote evaluation by a subscript.
A: 
Why do we write $T_h$ as the partial derivative w.r.t $h$ and what does it actually mean to take the derivative of the left action as we are doing in this definition?

$T_h$ is not at all a partial derivative with respect to $h$; note that in the setting of general manifolds, you cannot use the naive difference-quotient definition of derivatives, because in general, you cannot subtract points on a manifold. 
I suggest you take a look at Chapter $4$ on manifolds, in particular where they define tangent vectors. The authors define the tangent mapping between manifolds as follows: Given a smooth map $f: M \to N$ between two smooth manifolds, and a point $m \in M$, the tangent mapping at $m$ is denoted by $T_mf: T_mM \to T_{f(m)}N$, and is a linear map between tangent spaces defined by the rule
\begin{align}
(T_mf)([c]) := [f \circ c]
\end{align}
In words, it takes an equivalence class of curves $[c]$ in $T_mM$ and sends it to the equivalence class of curves $[f \circ c]$ in $T_{f(m)}N$ (recall that they define a tangent vector to be an equivalence class of smooth curves).
So, in your context, the notation $T_h$ does not mean partial derivatives at all. The symbol $T_h(L_g)$ means the tangent mapping at $h$ of the left-translation $L_g$, in the sense I described above. So, the way to "read" the formula
\begin{align}
(T_hL_g)X(h) = X(gh)
\end{align}
is as follows: first of all, $X$ is a vector field on the Lie group $G$, which means at each point $p\in G$, $X(p)$ is a tangent vector in the tangent space $T_pG$. Next, left-translation by $g$ is a smooth map $L_g:G \to G$, and we are taking its tangent at the point $h$. So, $T_h(L_g)$ is a linear mapping of $T_hG \to T_{L_g(h)}G = T_{gh}G$, hence $T_h(L_g)$ can be evaluated on the tangent vector $X(h)$. So, if we use a lot of brackets, we would write this as:
\begin{align}
\big(T_h(L_g) \big)(X(h))
\end{align} 
but omitting some brackets, we just write $(T_hL_g)X(h)$, and in words it says "the tangent mapping at $h$ of the smooth map $L_g$, evaluated on the tangent vector $X(h)$". The left-invariance condition of a vector field requires that all of this be equal to the tangent vector $X(gh)$.

Like you observed, there is a lot of different notation (and terminology) around; some authors use $T_mf$ to mean the tangent mapping at $m$, while other authors (e.g Lee) use $df_m$ to mean the linear map from $T_mM \to T_{f(m)}N$. Another thing you may notice is that the actual construction of tangent space varies from author to author, this can be confusing but for starters, the key thing you should focus on is what the formulas are really trying to say geometrically.
