Clearly the statement is true if $U$ is symmetric (,i.e. $U=U^{-1}$). Then we can pick $W$ as a symmetric subentourage of $V$ and then $U\circ W=W\circ U \subset V\subset U$ holds. So the only chance is to pick a nonsymmetric $U$.

 The second statement follows from the first one by passing to inverses (i.e. reflecting along the diagonal).

Now consider the uniform structure on $\mathbb{R}$ consisting of all subsets of $\mathbb{R}^2$ that contain an open neighborhood of the diagonal. 

Just to avoid confusion with left and right, I want to use the notation
$U\circ V = \{(x,z)| \exists y: (x,y)\in U \wedge (y,z)\in V\}.$

Let $U:=\{(x,y)| |y-x|<1 \}\cup  \mathbb{R}\times \{0\}$.
$V:=\{(x,y)| |y-x|<1 \}$. 
Then $V\circ U=\{(x,y)| |y-x|<2 \}\cup \mathbb{R}\times \{0\}$. 

Now let $W$ be any entourage and let $Z$ be some open neighborhood of $0$ such that $\{0\}\times Z\subset  W$. Then $U\circ W$ will contain $\mathbb{R}\times Z$. Thus $U\circ W$ will not be contained in $V\circ U$. Since $W$ was arbitrary this should be a counterexample.