How can I prove that the suspension of an Anosov diffeomorphism is an Anosov flow? Suppose we have an Anosov diffeo $f$ on $M$. Define $g_t : M\times\mathbb R\to M\times\mathbb R$ by $(x,s)\mapsto (x,s+t)$. Take a quotient of $M\times\mathbb R$ under the relation $(x,s)\sim (f(x),s-1)$. The flow descends to a flow $f_t$ on the quotient manifold. 
My question is : As the flow is locally a translation, how can it contract the stable direction? I see that a contraction does take place at regular intervals but is that enough?
 A: The short answer, as pointed out in the comments, is that yes, contraction at regular intervals is enough, because the definition of Anosov requires that there are $C>0$ and $\lambda\in (0,1)$ such that $\|Df^n|_{E^s}\| \leq C \lambda^n$ for all $n\geq 0$, and similarly for $\|Df^{-n}|_{E^u}\|$.  The constant $C$ allows the possibility that there are some points at which $E^s$ does not contract and/or $E^u$ does not expand under one iterate (or even several iterates) of $f$.  However, if you take $n$ large enough that $C\lambda^n < 1$, then $n$ iterates is always enough to see the hyperbolicity, no matter where you start.
There's a subtlety in your question that is worth pointing out, though.  The definition of Anosov involves a Riemannian metric (although for a compact manifold the property of being Anosov is in fact independent of the choice of metric).  If we write $M_1$ for the quotient manifold, then in order to ask "is the flow $f_t$ Anosov on $M_1$?" we need to first choose a Riemannian metric to use on $M_1$.  Even though $M_1$ is a quotient of $M\times \mathbb{R}$, which carries a natural Riemannian metric, you can't simply push the metric down under the quotient map $\pi \colon M\times \mathbb{R}\to M_1$, because $f$ is not an isometry, so pushing the metric from $y\in \pi^{-1}(x) \subset M\times \mathbb{R}$ to $x\in M_1$ may give a different result depending on which point $y\in \pi^{-1}(x)$ you choose.  And if you try to make a single global choice, say by identifying $M_1$ with $M\times [0,1)$ so that $\pi$ becomes a bijection (which I suspect is the picture in your head when you say that the flow is "locally a translation"), then you wind up with a discontinuity at the points where you glue $(x,1)$ to $(f(x),0)$.
Thus it's not immediately obvious what Riemannian metric to use.  One way to proceed is to write $h_0 \colon TM\times TM \to \mathbb{R}$ for the original metric on $M$, put $h_1 = h_0 \circ Df$, connect $h_0$ to $h_1$ by a family of metrics $h_t$ that varies smoothly in $t$, and then extend periodically to $\{h_t \}_{t\in\mathbb{R}}$, so that $h_{t+n} = h_t \circ Df^n$ for all $n\in\mathbb{Z}$.  Using $h_t$ as the metric on $M\times \{t\}$ and declaring $T_{(x,t)}(M\times\{t\})$ to be orthogonal to $T_{(x,t)}(\{x\}\times \mathbb{R})$ gives a metric on $M\times\mathbb{R}$ on which $(x,t) \mapsto (f(x),t-1)$ acts isometrically, so this metric does descend to a metric on $M_1$.  Then you can check that the Anosov condition holds.
