Your method certainly works, because you are just identifying boundary edges of the annulus $[0,1] \times S^1$ in pairs to form a surface. As usual, when one glues up edge pairs of a surface-with-boundary, the endpoints of the boundary edges form "vertex cycles" whose images are points of the quotient surface, and the smooth structure then extends across each of these points. This works regardless of rationality, irrationality, etc. of the lengths of the intervals.
I would describe the method differently, though, in order to isolate the "singular" behavior in a way that is easier to analyze. It is a somewhat standard method, and is more-or-less equivalent to your method.
To describe the standard method, let me set up notation for the interval exchange map. One partitions the circle $S^1$ into $M$ intervals in two ways, $S^1 = I_1 \cup \cdots \cup I_M$ and $J_1 \cup \cdots J_M$, and so that for each $m=1,\ldots,M$ the lengths of $I_m$ and $J_m$ are equal, and then one chooses an isometry $f_m : I_m \to J_m$. The map $\phi : S^1 \to S^1$ itself is then just the union of the maps $f_1,\ldots,f_m$, perhaps restricted to the interiors of the intervals $I_m$ in order to get something well-defined and continuous (or perhaps restricted to half-open intervals in some manner to get something bijective but discontinuous, something that never seems worthwhile to me).
To get a surface, one proceeds in a few steps.
First, let $X$ be the quotient space obtained from the disjoint union of $S^1$ and the $m$ rectangles $[0,1] \times I_m$, by identifying $(0,x)$ to $x$ and identifying $(1,x)$ to $f_m(x)$ for each $m$ and each $x \in I_m$. The lower corners of these rectangles are identified 2--1 with the endpoints of the $I$'s, and the upper corners are identified 2--1 with the endpoints of $J$'s. This object $X$ might be a surface-with-boundary, if you are lucky, but it won't be if there is a common endpoint of the $I$'s and the $J$'s.
Next one studies the "boundary cycles" of $X$ (the quotes are there to accomodate the possibility that $X$ is not actually a surface and these boundary cycles might touch where the $I$'s and $J$'s have a common endpoint). Construct each of these cycles as follows. Start with an $I$-endpoint $x_1 \in S^1$. Of the two lower corners identified with $x_1$, pick one, $p_1 \in [0,1] \times I_{m_1}$, say. Then go up a vertical side of $[0,1] \times I_{m_1}$ to an upper corner $p_2$, identified with some $J$-endpoint $y_1 \in S^1$. Then jump to the other upper corner identified with $y_1$, say $p_3 \in [0,1] \times I_{m_2}$. Then go down a vertical side of $[0,1] \times I_{m_2}$ to a lower corner $p_4$, identified with some $I$-endpoint $x_2$. Then jump to the other lower corner identified with $x_2$, say $p_5 \in [0,1] \times I_{m_3}$. Altogether, so far, you have travelled up a vertical side of $X$ from $x_1$ to $y_1$ and down another vertical side from $y_1$ to $x_2$. Now continue the cycle, which will go up-down-up-down... across a cycle of vertical sides that closes up to form a circle in $X$ subdivided into some even number of sides.
Finally, for each of these boundary cycles subdivided into $2m$ vertical sides, one subdivides each vertical side at its midpoint into two half length subarcs, and then: identify the two subarcs incident to $x_1$ to a single arc; identify the two subarcs incident to $y_1$ to a single arc; identify the two subarcs incident to $x_2$ to a single arc; and so on around the circle. The effect is to collapse the entire circle onto a $2m$-pointed star.
The quotient is a closed surface $S$. The vertical segments $[0,1] \times (point)$ glue up to form a foliation of the surface. This foliation has a $2m$-pronged singularity'' at the valence $2m$ star point associated to each boundary cycle consisting of $2m$ vertical sides. This foliation has an invariant transverse measure, coming from the $dx$-measure on each rectangle $[0,1] \times I_m$, and the fact that each gluing map $f_m$ was an isometry.
One can proceed to a smooth structure on $S$ as follows. The smooth structures on the rectangles $[0,1] \times I_m$ glue up to form a smooth structure everywhere except at the finite set of even-pronged singularities. Moreover, the Euclidean structures on these rectangles glue up to form a Euclidean metric except at the singularities. From the Euclidean metric, you get a conformal structure everywhere except at the prong singularities. But each prong singularity is removable, so the conformal structure is extendable across the prongs, which induces an extension of the smooth structure as well.
