Let $d \geq 1$ and let $V$ be a $2d+1$ irreducible representation of $SL(2,\mathbb{R})$. We know the irreducible representations of $SO(2)$ are the $2$ dimensional spaces $V_k$ with the map $$\rho_k : \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \mapsto \begin{pmatrix} \cos k\theta & -\sin k\theta \\ \sin k\theta & \cos k\theta \end{pmatrix}$$
When we restrict $V$ to $SO(2)$, we can write $V = V_0 + V_{k_1} + \cdots + V_{k_d}$. How can we figure out what $k_1,\ldots,k_d$ are?
I have reason to believe that $k_i = i$ for each $i$. I am basically trying to figure out why this is the case.
I assume you talk about real representations.
Realize $V$ as homogeneous polynomials of dergee $2d$ in $x,y$ and let $z=x+iy$, $\bar{z}=x-iy$. Homogenous polynomials are generated by real and imaginary parts of $z^k \bar{z}^{2d-k}$, $k=0,\ldots,d$. The action of $\varphi$-rotation is $$ z^k \bar{z}^{2d-k}\mapsto (e^{i\varphi k} z^k) (e^{-i\varphi (2d-k)} \bar{z}^{2d-k})=e^{i\varphi (2k-2d)} z^k \bar{z}^{2d-k}. $$ This says that $z^k \bar{z}^{2d-k}$ is mapped to a (complex) multiple of itself. Similarly, the conjugate $\bar{z}^k z^{2d-k}$ is invariant and it follows that the real and imaginary parts of $z^k \bar{z}^{n-k}$ generate a $2$-dimensional $SO(2)$-submodule. The case $k=d$ is special, as $z^{d} \bar{z}^{d}=(x^2+y^2)^{d}$ is purely real and clearly fixed by $SO(2)$: this corresponds to your $V_0$ copy.
The equation $$R(\varphi) z^k \bar{z}^{2d-k}=e^{i\varphi (2k-2d)} z^k \bar{z}^{2d-k} $$ shows that in the basis $(Re(z^k \bar{z}^{2d-k}), Im(z^k \bar{z}^{2d-k}))$ the rotation actions can be identified with your $\rho_{2k-2d}$. The remaining numbers $2k-2d$ for $k=0,\ldots, d-1$ equal $\{-2d,2-2d,\ldots, -2\}$ and these are your constants $k_1,\ldots, k_d$. They could also be $2,4,\ldots,2d$: the numbers $k_i$ are well-defined only up to the sign. The sign of $k_i$ depends on the choice of the orientation of the $2$-dimensional subspaces.
For an explicit example, consider $d=1$ and $V=\langle x^2, xy, y^2\rangle$ with the obvious action. Then $V_0=\langle x^2+y^2 \rangle$ and $V_{\pm 2}=\langle Re(z^2), Im(z^2)\rangle = \langle x^2-y^2, 2xy \rangle$. Choosing the basis $(x^2-y^2, 2xy)$, the rotation acts like your $\rho_2$.
Remark 1: If you don't have an orientation, you don't know whether $\rho_k(\varphi)(p)$ moves clockwise or counterclockwise for $\varphi\in [0,2\pi]$, $p$ a point in the plane. Note also that if you choose some base of your $2$-dimensional invariant space, the $SO(2)$-action doesn't have to be represented by orthogonal matrices at all. But $|k_j|$ is well-defined.