Does a Lie derivative of a vector field involve subtracting vectors from different spaces? Addition (and subtraction) is not by default defined for vectors in different spaces, even if those vector spaces are isomorphic (it is possible to define addition, but there are many ways to define it). Addition of vectors in different tangent spaces can be defined if a local chart is given (the natural thing to do is to identify basis vectors with each other), but depending on the chart, the addition or subtraction can be different.
That said, a Lie derivative can be defined as,
$$
\lim_{t\rightarrow 0} \frac{\phi_t[Y] - Y}{t}
$$
where $\phi_t$ is the $t$th transformation in a one-parameter group, and $Y$ is a vector field. As far as I can tell, for a point $x$, $\phi_t[Y](x)$ and $Y(x)$ are vectors inhabiting different spaces. So, how are we justified in subtracting them? Perhaps the answer has something to do with the fact that the spaces are "infinitesimally close," but that would require more development to be truly satisfying.
 A: The answer to your title question is NO, everything is happening in the same vector space. The definition given is incorrect (or at the very least not using "standard" notation). We have to first describe the notion of "pulling back" a vector field by a diffeomorphism:

Definition.
Let $M,N$ be smooth manifolds, $\phi:M \to N$ a diffeomorphism, and $Y$ a smooth vector field on $N$. We define the pull-back vector field $\phi^*[Y]$ (also written $\phi^*Y$ or with any other bracketing convention you like) by
\begin{align}
\phi^*[Y] := T \phi^{-1} \circ Y \circ \phi
\end{align}
where in general for a map $g:M \to N$, $Tg:TM \to TN$ refers to the tangent map.

Note of course that in the above definition, in writing $T \phi^{-1}$, it doesn't matter if you interpret it as $T(\phi^{-1}):TN \to TM$ or $(T \phi)^{-1}:TN \to TM$, because it's the same thing.
Again, the idea of pulling back is that we have a vector field $Y$ on the target manifold $N$, and we would like to have a vector field on the domain manifold. So, what's the most natural thing to try? Well, take a point $p \in M$ in the domain. We somehow need to get a vector in $T_pM$. To do this, we first send $p$ to $\phi(p) \in N$, then use the vector field $Y$ to get a vector $Y(\phi(p)) \in T_{\phi(p)}N$, and then finally we "bring back" this vector using $T\phi^{-1}$ to get the vector $T\phi^{-1}(Y(\phi(p))) \in T_{\phi^{-1}(\phi(p))}M = T_pM$. Then, finally since all the maps are smooth, it follows that $\phi^*[Y]$ is also a smooth vector field.

Now, in the definition of the Lie derivative, we have the same manifolds, $M=N$, and the diffeomorphism in question is $\phi_t:M \to M$, the time $t$ flow map. So, per the above construction, we hve a vector field $Y$ on "the target manifold" $M$, and we pull it back to get a vector field on the "domain manifold" $M$, via $(\phi_t)^*[Y]$. So, all the operations are taking place in the same vector space: for each $p \in M$, we have $[(\phi_t)^*Y](p) \in T_pM$ and $Y(p) \in T_pM$ also (this is of course true, because they're both vector fields on $M$, so if I evaluate at $p \in M$, I get vectors in $T_pM$, so the subtraction and division by $t$ is also happening in $T_pM$).
