Can a periodically additively perturbed sinusoidal vector field on the circle have a stable periodic orbit of higher least period?

The claim is true. (As proved here, it is quite unique to sinusoidal vector fields. The same reference also mentions, in its introduction, existing applications of equation (1) with $g$ taking the particular form mentioned above.)

In short (as in Theorem 1 of here): the stereographic projections of the solutions of (1) satisfy a Riccati equation (1st order quadratic ODE), and (as in Proposition 2 of here) the time-$t$ mappings of Riccati equations are Möbius transformations; the classification of Möbius transformations then gives the result. In fact, either: [A] all solutions of (1) are neutrally stable; or [B] (1) has a unique asymptotically stable $1$-periodic solution and a unique unstable $1$-periodic solution, and all solutions other than the unstable $1$-periodic solution are attracted to the stable $1$-periodic solution; or [C] (1) has a unique $1$-periodic solution, and this solution attracts all solutions but is unstable.

To go through the proof in full detail:

Lemma 1. For any solution $\theta(\cdot)$ of (1), letting $y(\cdot)$ be the stereographic projection $y(t)=\tan(\frac{\theta(t)}{2})$, we have that $$ \dot{y}(t) \ = \ Ay(t) \ + \ \tfrac{1}{2}g(t)(1+y(t)^2) $$ whenever $\theta(t) \neq \pi$.

Proof. Straightforward direct computation. $\ \square$

Now let $\mathbb{S}_4$ be the set of points in $(\mathbb{S}^1)^4$ whose four coordinates are distinct. Define $C \colon \mathbb{S}_4 \to \mathbb{R}$ to be the cross-ratio of the stereographic projections $\tan(\frac{\cdot}{2})$ of the four inputs; note that $C$ is a smooth function, as it can be written explicitly (i.e. without reference to arithmetic with infinity) as $$ C(\theta_1,\theta_2,\theta_3,\theta_4) \ = \ \left\{ \begin{array}{c l} \frac{\left(\tan(\frac{\theta_3}{2}) \tan(\frac{\pi-\theta_1}{2})-1\right)\left(\tan(\frac{\theta_4}{2}) - \tan(\frac{\theta_2}{2})\right)}{\left(\tan(\frac{\theta_3}{2}) - \tan(\frac{\theta_2}{2})\right)\left(\tan(\frac{\theta_4}{2}) \tan(\frac{\pi-\theta_1}{2})-1\right)} & \theta_1 \neq 0 \textrm{ and } \theta_2,\theta_3,\theta_4 \neq \pi \\ & \\ \frac{\left(\tan(\frac{\theta_3}{2}) - \tan(\frac{\theta_1}{2})\right)\left(\tan(\frac{\theta_4}{2}) \tan(\frac{\pi-\theta_2}{2})-1\right)}{\left(\tan(\frac{\theta_3}{2}) \tan(\frac{\pi-\theta_2}{2})-1\right)\left(\tan(\frac{\theta_4}{2}) - \tan(\frac{\theta_1}{2})\right)} & \theta_2 \neq 0 \textrm{ and } \theta_1,\theta_3,\theta_4 \neq \pi \\ & \\ \frac{\left(1-\tan(\frac{\pi-\theta_3}{2}) \tan(\frac{\theta_1}{2})\right)\left(\tan(\frac{\theta_4}{2}) - \tan(\frac{\theta_2}{2})\right)}{\left(1-\tan(\frac{\pi-\theta_3}{2}) \tan(\frac{\theta_2}{2})\right)\left(\tan(\frac{\theta_4}{2}) - \tan(\frac{\theta_1}{2})\right)} & \theta_3 \neq 0 \textrm{ and } \theta_1,\theta_2,\theta_4 \neq \pi \\ & \\ \frac{\left(\tan(\frac{\theta_3}{2}) - \tan(\frac{\theta_1}{2})\right)\left(1-\tan(\frac{\pi-\theta_4}{2}) \tan(\frac{\theta_2}{2})\right)}{\left(\tan(\frac{\theta_3}{2}) - \tan(\frac{\theta_2}{2})\right)\left(1-\tan(\frac{\pi-\theta_4}{2}) \tan(\frac{\theta_1}{2})\right)} & \theta_4 \neq 0 \textrm{ and } \theta_2,\theta_3,\theta_4 \neq \pi. \end{array} \right. $$

Lemma 2. $C(\cdot)$ is a conserved quantity for the four-point motion of (1).

For this, we need the following very simple fact:

Lemma 3. (A) For any linear map $L \colon \mathbb{R} \to \mathbb{R}$ and quantities $y_1,y_2,y_3 \in \mathbb{R}$, $$ (y_2-y_1)(L(y_3)-L(y_1)) \ = \ (L(y_2)-L(y_1))(y_3-y_1). $$ (B) For any quadratic map $Q \colon \mathbb{R} \to \mathbb{R}$ and quantities $y_1,y_2,y_3,y_4 \in \mathbb{R}$, \begin{align*} & (y_3-y_1)(y_4-y_2)[(y_3-y_2)(Q(y_4)-Q(y_1)) + (Q(y_3)-Q(y_2))(y_4-y_1)] \\ = \ & [(y_3-y_1)(Q(y_4)-Q(y_2)) + (Q(y_3)-Q(y_1))(y_4-y_2)](y_3-y_2)(y_4-y_1). \end{align*}

Proof. (A) Writing $L(x)=ax$, we have that $$ \mathrm{LHS} \ = \ a(y_2-y_1)(y_3-y_1) \ = \ \mathrm{RHS}. $$ (B) Writing $Q(x)=ax^2+bx+c$, we have that \begin{align*} \mathrm{LHS} \ &= \ 2b(y_3-y_1)(y_4-y_2)(y_3-y_2)(y_4-y_1) \\ & \hspace{10mm} + a(y_3-y_1)(y_4-y_2)(y_3-y_2)(y_4^2-y_1^2) \\ & \hspace{10mm} + a(y_3-y_1)(y_4-y_2)(y_4-y_1)(y_3^2-y_2^2) \\ &= \ (y_3-y_1)(y_4-y_2)(y_3-y_2)(y_4-y_1)[2b + a(y_1+y_2+y_3+y_4)]. \end{align*} One obtains the same for the $\mathrm{RHS}$. $\ \square$

Proof of Lemma 2. Let $\theta_1(t),\theta_2(t),\theta_3(t),\theta_4(t)$ be four distinct solutions of (1), with projections $y_1(t),y_2(t),y_3(t),y_4(t)$ respectively, and let $C_0(t)=C(\theta_1(t),\theta_2(t),\theta_3(t),\theta_4(t))$. We need that for all $t$, $\dot{C}_0(t)=0$. Firstly, if $U \subset \mathbb{R}$ is an open time-interval during which $\theta_i(t)$ is fixed at $\pi$ for some $i$, then $g(t)=0$ on $U$ (since $\sin(\pi)=0$) and $$ C_0(t) \ = \ \pm\frac{y_j(t)-y_k(t)}{y_j(t)-y_l(t)} $$ on $U$; so one can use Lemma 1 (with $g(t)=0$) to compute $\dot{C}_0(t)$, and due to Lemma 3(A) one obtains that $\dot{C}_0(t)=0$. Secondly, if we have a time $t$ at which $\theta_1(t),\theta_2(t),\theta_3(t),\theta_4(t) \neq \pi$, then $$ C_0(t) \ = \ \frac{(y_3(t)-y_1(t))(y_4(t)-y_2(t))}{(y_3(t)-y_2(t))(y_4(t)-y_1(t))} \, ; $$ so one can use Lemma 1 to compute $\dot{C}_0(t)$, and due to Lemma 3(B) one obtains that $\dot{C}_0(t)=0$. Now the solutions $\theta_i(t)$ are continuously differentiable (since $g$ is continuous), and so since $C$ is smooth, we have that $C_0(\cdot)$ is continuously differentiable. Thus we can conclude overall that $\dot{C}_0(t)=0$ for all $t$. $\ \square$

Now let $\hat{\mathbb{C}}=\mathbb{C} \cup \{\infty\}$ be the one-point compactification of $\mathbb{C}$, and let $\hat{\mathbb{R}}=\mathbb{R} \cup \{\infty\} \subset \hat{\mathbb{C}}$. A Möbius transformation is a map $T \colon \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ taking either the form $T(z)=az+b$ with $a \neq 0$ or the form $T(z)=\frac{a}{z+b}+c$ with $a \neq 0$.

Lemma 4. Fix $t \geq 0$, and let $f \colon \mathbb{S}^1 \to \mathbb{S}^1$ be the time-$t$ mapping of (1). Let $F \colon \hat{\mathbb{R}} \to \hat{\mathbb{R}}$ be the conjugation of $f$ by the projection $\tan(\frac{\cdot}{2})$. Then $F$ is the restriction to $\hat{\mathbb{R}}$ of a Möbius transformation.

Proof. Pick any distinct points $y_1,y_2,y_3 \in \mathbb{R} \cap F^{-1}(\mathbb{R})$. For every $x \in (\mathbb{R} \setminus \{y_1,y_2,y_3\} ) \cap F^{-1}(\mathbb{R})$, Lemma 2 gives that $$ \frac{(F(y_3)-F(y_1))(F(x)-F(y_2))}{(F(y_3)-F(y_2))(F(x)-F(y_1))} \ = \ \frac{(y_3-y_1)(x-y_2)}{(y_3-y_2)(x-y_1)}. $$ It is easy to see that one can rearrange the above equation to make $F(x)$ the subject, with the right-hand side being expressible as the ratio of two affine functions of $x$. Hence, since $F$ is continuous and injective, $F$ is a Möbius transformation (restricted to $\hat{\mathbb{R}}$). $\ \square$

Now it is known (p37 of here) that any Möbius transformation $T$ is conjugate to either the translation $z \mapsto z+1$ or a linear map $z \mapsto \lambda z$ for some $\lambda \in \mathbb{C} \setminus \{0\}$. Fixing $t \geq 0$ and applying this to the Möbius transformation $T$ whose restriction to $\hat{\mathbb{R}}$ coincides with $F$ (as defined in Lemma 4):

• If $h \circ T \circ h^{-1}\colon z \mapsto z+1$, then the topological circle $h(\hat{\mathbb{R}})$ is strictly invariant under $z \mapsto z+1$ and so contains $\infty$; hence $h^{-1}(\infty)$ lies in $\hat{\mathbb{R}}$ and is the unique fixed point of $F$.
• If $h \circ T \circ h^{-1}\colon z \mapsto \lambda z\,$ where $|\lambda| \neq 1$, then the topological circle $h(\hat{\mathbb{R}})$ contains both $0$ and $\infty$; hence $h^{-1}(0)$ and $h^{-1}(\infty)$ are the fixed points of $F$. If $|\lambda|>1$ then all trajectories of $F$ not starting at $h^{-1}(0)$ are attracted to $h^{-1}(\infty)$; if $|\lambda|<1$ then all trajectories of $F$ not starting at $h^{-1}(\infty)$ are attracted to $h^{-1}(0)$.
• If $h \circ T \circ h^{-1}\colon z \mapsto e^{i\mu} z\,$ where $\mu \in \mathbb{R} \setminus \{0\}$, then $h(\hat{\mathbb{R}})$ is a circle of non-zero finite radius about the origin, and so $F$ is topologically conjugate to the circle rotation by angle $\mu$.
• Finally, if $h \circ T \circ h^{-1}$ is the identity function then $F$ is the identity function.

Since $F$ is conjugate to the time-$t$ map $f$ of (1), this completes the proof of the result.