In Geometrical Methods for Mathematical Physics, Schutz speaks of a left-invariant vector field $\bar{V}$ at the identity $e$ of a Lie group $G$. He writes:
Unlike an arbitrary vector field, $\bar{V}$ is determined completely by $\bar{V}_e$, so we can denote the points of $G$ on this curve by $$g_{\bar{V}_e}(t)=\exp(t\bar{V}_e).$$ Because exponentiation has, by definition, the property $$\exp(t_2\bar{V})\exp(t_1\bar{V})|_e=\exp\left[(t_1+t_2)\bar{V}\right]|_e,$$ the points on these integral curves form a group ...
It is my understanding, however, that exponentiation does not necessarily have the sum-to-product property when dealing with vectors/matrices. What is the deeper reason behind Schutz making this claim?
Well it is true that $\exp(X)\exp(Y)$ in general is not
$\exp(X+Y)$. In fact you have a very well known formula that is the Baker-Campbell-Hausdorff Formula that states $$\exp\left(X\right)\exp\left(Y\right)=\exp\left(X+Y+\frac{1}{2}\left[X,Y\right]+\frac{1}{12}\left(\left[X,\left[X,Y\right]\right]+\left[Y,\left[Y,X\right]\right]\right)...\right)$$ As you can see since the brackets are linear and $\left[X,X\right]=0 $you have $$\exp\left(t_1X\right)\exp\left(t_2X\right)=\exp\left(t_1X+t_2X+\frac{t_1t_2}{2}\left[X,X\right]+...\right)=\exp\left(t_1X+t_2X\right)$$ Another way of seeing the same thing in this particular case is just writing the series of the exponential and grouping $t_1,t_2$.