On uniparametric subgroups of a Lie group Suppose $x,y$ are elements in some Lie algebra $\mathfrak{g}$. Using the Taylor expansion for $\mathrm{exp}$ we know that $\mathrm{exp}(tx)\mathrm{exp}(ty)=\mathrm{exp}(t(x+y)+\frac{t^2}2[x,y]+\dots)$. If $[x,y]$ vanishes then both $\mathrm{exp}(tx)\mathrm{exp}(ty)$ and $\mathrm{exp}(ty)\mathrm{exp}(tx)$ are curves that pass through $e$ at $t=0$ with derivative $x+y$, so by a uniqueness theorem in ODEs they agree everywhere (and in fact define the uniparametric group $\mathrm{exp}(t(x+y))$). EDIT: This is wrong, the theorem in ODEs says that two integral curves for the same vector field agreeing in one point agree everywhere. In the case that the bracket vanishes, these two curves are integral curves for the left-invariant field corresponding to the vector $x+y$.
Suppose now that $[x,y]$ does not vanish. Then obviously $\mathrm{exp}(tx)\mathrm{exp}(ty)$ does not agree with $\mathrm{exp}(ty)\mathrm{exp}(tx)$, but still the first is a smooth curve that passes through $e$ at $t=0$ with derivative $x+y$, so why isn't it the uniparametric group $\mathrm{exp}(t(x+y))$?
 A: Because then $t\mapsto\exp(tx)\exp(ty)$ is not a group homomorphism from $\mathbb R$ into $G$.
A: Your question is difficult to answer: that curve is not a uniparametric subgroup simply because it isn't… I suggest you consider examples. Easy ones come from taking $\mathfrak{g}=\mathfrak{gl}_n$ for some $n$ and $x$ and $y$ strictly upper triangular, so that the exponentials are easy to compute explicitly.
Take for the smallest example, $x=\left(
\begin{array}{ccc}
 0 & 1 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right)$ and $y=\left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 0 & 1 \\
 0 & 0 & 0 \\
\end{array}
\right)$. Then 
\begin{gather}\exp(tx)=\left(
\begin{array}{ccc}
 1 & t & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right)\\
\exp(ty)=\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & t \\
 0 & 0 & 1 \\
\end{array}
\right)\\
\exp(tx)\exp(ty)=\left(
\begin{array}{ccc}
 1 & t & t^2 \\
 0 & 1 & t \\
 0 & 0 & 1 \\
\end{array}
\right)\\
\exp(t(x+y))=\left(
\begin{array}{ccc}
 1 & t & \frac{t^2}{2} \\
 0 & 1 & t \\
 0 & 0 & 1 \\
\end{array}
\right).
\end{gather}
In particular,
$$\exp(tx)\exp(ty)\cdot \exp(sx)\exp(sy)=\left(
\begin{array}{ccc}
 1 & s+t & s (s+t)+t^2 \\
 0 & 1 & s+t \\
 0 & 0 & 1 \\
\end{array}
\right)$$ while $$\exp((s+t)x)\exp((s+t)y)=\left(
\begin{array}{ccc}
 1 & s+t & (s+t)^2 \\
 0 & 1 & s+t \\
 0 & 0 & 1 \\
\end{array}
\right).$$
