There is no hope of comparisons between $\pi_0\operatorname{Hom}^L_X(x,y)$ and $\pi_0\operatorname{Hom}^R_X(x,y)$ in general, even when $x=y$.

Example. Let $X$ be the coequaliser of $f,g \colon \Delta^0 \amalg \Delta^1 \rightrightarrows \Delta^2$, where $f$ maps $\Delta_0$ to $0$ and $\Delta^1$ to $\operatorname{id}_0$, and $g$ maps $\Delta_0$ to $2$ and $\Delta^1$ to the arrow $0 \to 1$. The nondegenerate simplices in $X$ are:

• A single vertex $x$ (the image of $0$, $1$, and $2$ in $\Delta^2$).
• Two arrows $\alpha$ and $\beta$ (the images of $0 \to 2$ and $1 \to 2$; the third arrow $0 \to 1$ has become identified with a degenerate $1$-simplex).
• One $2$-simplex $h$ (the image of $0 \to 1 \to 2$), by definition a right homotopy $\alpha \stackrel\sim\to \beta$.

Thus we see that $\alpha$ and $\beta$ are right homotopic (hence homotopic in $\operatorname{Hom}_X(x,x)$ as well), but not left homotopic as there are no other nondegenerate $2$-simplices. $\square$

In particular, any of the definitions that are self-dual (e.g. $\operatorname{Hom}_{X^{\text{op}}}(x,y) = \operatorname{Hom}_X(y,x)^{\text{op}}$) cannot agree with $\pi_0\operatorname{Hom}^L_X(x,y)$ or $\pi_0\operatorname{Hom}^R_X(x,y)$ in general. For instance, if $\pi_0\operatorname{Hom}^L_X(x,y) = \pi_0\operatorname{Hom}_X(x,y)$ for all $X$, then \begin{align*} \pi_0\operatorname{Hom}_X^R(x,y) &= \pi_0\operatorname{Hom}_{X^{\text{op}}}^L(y,x)^{\text{op}} = \pi_0 \operatorname{Hom}_{X^{\text{op}}}(y,x)^{\text{op}} \\ &= \pi_0 \operatorname{Hom}_X(x,y) = \pi_0 \operatorname{Hom}_X^L(x,y), \end{align*} which we saw is false (again feel free to assume $x=y$).

I believe all the others are self-dual (but I'm not familiar with all of them), so at least they don't agree with $\pi_0\operatorname{Hom}_X^L(x,y)$ or its dual. I'm not sure if any subset of the others agree in general, but you can probably make counterexamples like the one above.

There is no hope of comparisons between $\pi_0\operatorname{Hom}^L_X(x,y)$ and $\pi_0\operatorname{Hom}^R_X(x,y)$ in general, even when $x=y$. (Note that the statement you cited says $\operatorname{Hom}^L_{X^{\text{op}}}(x,x) \cong \operatorname{Hom}_X^R(x,x)^{\text{op}}$, so it computes Homs in a different simplicial set.)

Example. Let $X$ be the coequaliser of $f,g \colon \Delta^0 \amalg \Delta^1 \rightrightarrows \Delta^2$, where $f$ maps $\Delta_0$ to $0$ and $\Delta^1$ to $\operatorname{id}_0$, and $g$ maps $\Delta_0$ to $2$ and $\Delta^1$ to the arrow $0 \to 1$. The nondegenerate simplices in $X$ are:

• A single vertex $x$ (the image of $0$, $1$, and $2$ in $\Delta^2$).
• Two arrows $\alpha$ and $\beta$ (the images of $0 \to 2$ and $1 \to 2$; the third arrow $0 \to 1$ has become identified with a degenerate $1$-simplex).
• One $2$-simplex $h$ (the image of $0 \to 1 \to 2$), by definition a right homotopy $\alpha \stackrel\sim\to \beta$.

Thus we see that $\alpha$ and $\beta$ are right homotopic (hence homotopic in $\operatorname{Hom}_X(x,x)$ as well), but not left homotopic as there are no other nondegenerate $2$-simplices. $\square$

In particular, any of the definitions that are self-dual (e.g. $\operatorname{Hom}_{X^{\text{op}}}(x,y) = \operatorname{Hom}_X(y,x)^{\text{op}}$) cannot agree with $\pi_0\operatorname{Hom}^L_X(x,y)$ or $\pi_0\operatorname{Hom}^R_X(x,y)$ in general. For instance, if $\pi_0\operatorname{Hom}^L_X(x,y) = \pi_0\operatorname{Hom}_X(x,y)$ for all $X$, then \begin{align*} \pi_0\operatorname{Hom}_X^R(x,y) &= \pi_0\operatorname{Hom}_{X^{\text{op}}}^L(y,x)^{\text{op}} = \pi_0 \operatorname{Hom}_{X^{\text{op}}}(y,x)^{\text{op}} \\ &= \pi_0 \operatorname{Hom}_X(x,y) = \pi_0 \operatorname{Hom}_X^L(x,y), \end{align*} which we saw is false (again feel free to assume $x=y$).

I believe all the others are self-dual (but I'm not familiar with all of them), so at least they don't agree with $\pi_0\operatorname{Hom}_X^L(x,y)$ or its dual. I'm not sure if any subset of the others agree in general, but you can probably make counterexamples like the one above. (The convenient coincidence is that finite simplicial sets are far removed from the Kan or inner horn conditions: from some dimension on, you can only produce degenerate simplices, which do a bad job filling horns.)