Let $f=(x_1,x_2,\ldots,x_r)$. We show that $o(f)=r$. We claim that $f^r(x_i)=x_{(i+r)\mod{r}}$.
First, define $x_r = x_0$. Base case: $f^r(x_1) = x_{(1+r)\mod{r}}=x_1$.
Inductive step: show $f^r(x_i)=x_{(i+r)\mod{r}} \implies f^r(x_{i+1})=x_{(i+1+r)\mod{r}}$.
The reason I defined $x_r = x_0$ was to be able to right $i+1$ without worrying about the case where $i=r$.
Is this the proper set up? I can definitely show the inductive step. However, I am unsure how to justify my base case rigorously. Obviously, applying the function to an element r times should correspond to a "shift" by r. I could spell it out for $x_1$ by showing many steps:
$f(x_1) = x_2$
$f(x_2)=f^2(x_1)=x_3$
$f(x_3) = f^2(x_2) = f^3(x_1) = x_4$
$\vdots$
$f^r(x_1)=x_{1+r}=x_{0+1}=x_1$
Does this make this part of the proof work? I would follow like this:
$f^r(x_i)=x_{(i+r)\mod{r}}$
Operating with $f$ on both sides:
$f^{r+1}(x_i)=f^n(f(x_i))=f^n(x_{i+1})=f(x_{(i+r)\mod{r}})= x_{(i+r)\mod{r}+1}=x_{(i+r+1)\mod{r}}$
So,
$f^n(x_{i+1})=x_{(i+r+1)\mod{r}}$
Obviously, there is more to this proof but I am only interested in knowing about this part, irrespective of its final utility in the proof.
UPDATE: This is the proof I have come up with given the responses so far
Let $f = (x_1,x_2,...,x_{r})$. We show that $o(f)=r$. First, define $x_r = x_0$. Then, letting $\bigoplus$ represent addition modulo $r$, we claim that $f^n(x_i) = x_{i\bigoplus n}$. $f(x_i) = x_{i+1}$, by definition. It is also clear that this is consistent with $f(x_i) = x_{i\bigoplus 1}$. So, to verify our claim, we need only show that
$f^n(x_i) = x_{i\bigoplus n} \implies f^{n+1}(x_i)=x_{i\bigoplus (n+1)}$.
Beginning with our assumption, we have
$f^n(x_i) = x_{i\bigoplus n}$.
Operating on both sides with $f$,
$f(f^n(x_i)) = f^{n+1}(x_i) = f(x_{i\bigoplus n})$.
By our assumption,
$f(x_{i\bigoplus n}) = x_{i\bigoplus n\bigoplus 1} = x_{i\bigoplus (n+1)}$.
So, we may conclude
$f^n(x_i) = x_{i\bigoplus n} \implies f^{n+1}(x_i)=x_{i\bigoplus (n+1)}$.
We are therefore justified in saying
$f^r(x_i) = x_{i}$
since $i\bigoplus r = i$ for $0<i<r$ and $0$ when $i=r$. Recall that we defined $x_r = x_0$. So, $f^r=e$, the identity, and $o(f) \leq r$. Assume for the sake of contradiction that $o(f) = m < r$. Then, by our formula, this implies that $m\bigoplus i = i$. In other words, $m+i = qr + i$ for some $q\in \mathbf{Z}$. Subtracting $i$ from both sides, we find $m=qr$. Since $q\in Z$, $m\geq r$, which contradicts our assumption that $m<r$. Therefore, $o(f) = r$.
The sketch of a proof I give below may seem like I'm bypassing your answer to take a different approach instead. But the proof below is precisely what happens in your proof at the vertical dots, where you "spell it out for $x_1$ by showing many steps". Replacing $x_1$ by $x_i$ in this argument gives you a proof of your original statement; the rest of your argument is irrelevant:
Instead of induction on $i$, try induction on $k$ to show that $f^k(x_i)=x_{i+k}$, where we take the index mod $r$. For the base case $k=1$ you need to show that $$f(x_i)=x_{i+1},$$ for all $i$. This identity should be clear or immediate from whatever definition you are using. Then for the induction step you should show that for every positive integer $k$: $$\text{If }\quad f^k(x_i)=x_{i+k}\quad\text{ then }\quad f^{k+1}(x_i)=x_{i+k+1}.$$ Again this should follow quickly from the definition of $f$. Then it follows, perhaps with a short argument, that $f^k=\operatorname{id}$ if and only if $k\equiv0\pmod{r}$, and hence that $o(f)=r$.