How to see that this space with two vertices and four edges is homotopy equivalent to $S^1 \vee S^1 \vee S^1$? I am having trouble seeing why this space has fundamental group $\mathbb Z * \mathbb Z * \mathbb Z$.
I have read that this space is homotopy equivalent to $S^1 \vee S^1 \vee S^1$, from which we can conclude the fundamental groups are isomorphic.
How can I visualize this homotopy equivalence?

 A: To see that the space in the figure, call it $X$, is homotopy equivalent to $S^1 \vee S^1 \vee S^1$, use the first criteria for homotopy equivalence from Hatcher, page 11: 

If $(X,A)$ is a CW pair consisting of a CW complex $X$ and a contractible subcomplex $A$, then the quotient map $X \to X/A$ is a homotopy equivalence.

The space in the figure, $X$, is a CW-complex, and the edge labeled $a$ is a subcomplex, call it $A$. So, $(X,A)$ is a CW pair. Also, the subcomplex $A$ is contractible. So, contracting $A$ to a point gives us a space, $X/A$, which is a wedge of three copies of $S^1$. Hence, by the criteria, $X$ and $X/A \cong S^1 \vee S^1 \vee S^1$ are homotopy equivalent.
A: Juan's answer is probably best. But another way to visualize this is as follows.
Let $Y$ denote the space obtained by deleting three small disjoint discs from a bigger disc $D^2$. Embed your space $X$ into $Y$. Then there is a deformation retraction of $Y$ onto $X$. Similarly, you can get a deformation retraction of $Y$ onto the bouquet of three circles. 
