The point is that we don't know what $iM$ means if $M\in \U0001d530\U0001d532(2)$ because $\U0001d530\U0001d532(2)$ is a real Lie algebra so we need to complexify it, but we do know what $i{R}_{*}M\in \U0001d524\U0001d529(V)$ means because $\U0001d524\U0001d529(V)$ is a complex Lie algebra and we don't to complexify it.
X and Y
From su(2) to sl(2,C)
Last time we saw that, given a complex representation $R:SU(2)\to GL(V)$ there is a weight space decomposition $V={\oplus}_{j=1}^{n}{V}_{j}$ such that with respect to this splitting, $R\left(\begin{array}{cc}\hfill {e}^{i\theta}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {e}^{i\theta}\hfill \end{array}\right)$ is the diagonal matrix $$\left(\begin{array}{ccc}\hfill {e}^{i{m}_{1}\theta}\hfill & \hfill \hfill & \hfill 0\hfill \\ \hfill \hfill & \hfill \mathrm{\ddots}\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill \hfill & \hfill {e}^{i{m}_{n}\theta}\hfill \end{array}\right)$$ where ${m}_{j}\in \mathbf{Z}$ are the weights of the representation.
At the level of Lie algebras, we have $$R\left(\mathrm{exp}\theta \left(\begin{array}{cc}\hfill i\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill i\hfill \end{array}\right)\right)=\mathrm{exp}\theta {R}_{*}\left(\begin{array}{cc}\hfill i\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill i\hfill \end{array}\right),$$ so $${R}_{*}\left(\begin{array}{cc}\hfill i\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill i\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill i{m}_{1}\theta \hfill & \hfill \hfill & \hfill 0\hfill \\ \hfill \hfill & \hfill \mathrm{\ddots}\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill \hfill & \hfill i{m}_{n}\hfill \end{array}\right).$$
In other words, we know where one of the matrices in $\U0001d530\U0001d532(2)$ goes under ${R}_{*}$ . But there are other matrices in $\U0001d530\U0001d532(2)$ : the general element is $\left(\begin{array}{cc}\hfill ix\hfill & \hfill y+iz\hfill \\ \hfill y+iz\hfill & \hfill ix\hfill \end{array}\right)$ ; where do these go?
Pick the basis $${\sigma}_{1}=\left(\begin{array}{cc}\hfill i\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill i\hfill \end{array}\right),{\sigma}_{2}=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right),{\sigma}_{3}=\left(\begin{array}{cc}\hfill 0\hfill & \hfill i\hfill \\ \hfill i\hfill & \hfill 0\hfill \end{array}\right)$$ for $\U0001d530\U0001d532(2)$ (corresponding to the $x$ , $y$ and $z$ directions). These (or closely related versions of them) are often called the Pauli matrices.
I've told you what ${R}_{*}{\sigma}_{1}$ is, and since ${R}_{*}$ is linear (determined by its values on a basis) it's enough to say what ${R}_{*}{\sigma}_{2}$ and ${R}_{*}{\sigma}_{3}$ are to determine ${R}_{*}$ completely. It turns out to be much easier to use: $$X=\frac{1}{2}({\sigma}_{2}i{\sigma}_{3})=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),Y=\frac{1}{2}({\sigma}_{2}+i{\sigma}_{3})=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right).$$
The problem is that these correspond to complex values of $y$ and $z$ , so they don't live in $\U0001d530\U0001d532(2)$ . Instead, they live in $\U0001d530\U0001d532(2)\otimes \mathbf{C}$ . In fact, $X$ and $Y$ live in $\U0001d530\U0001d529(2,\mathbf{C})$ and ${\sigma}_{1}$ is equal to $iH$ where $H=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$ . So it turns out that $\U0001d530\U0001d532(2)\otimes \mathbf{C}=\U0001d530\U0001d529(2,\mathbf{C})$ : anything inside $\U0001d530\U0001d529(2,\mathbf{C})$ can be written as a complex linear combination of ${\sigma}_{1},{\sigma}_{2},{\sigma}_{3}$ and anything inside $\U0001d530\U0001d532(2)\otimes \mathbf{C}$ can be written as a complex linear combination of $X$ , $Y$ , and $H$ .
If ${R}_{*}:\U0001d530\U0001d532(2)\to \U0001d524\U0001d529(V)$ is a representation of Lie algebras where $V$ is a complex vector space, then we get for free a representation ${R}_{*}^{\mathbf{C}}:\U0001d530\U0001d532(2)\otimes \mathbf{C}\to \U0001d524\U0001d529(V)$ , just by setting ${R}_{*}^{\mathbf{C}}({M}_{1}+i{M}_{2})={R}_{*}{M}_{1}+i{R}_{*}{M}_{2}$ .
How do X and Y act?
Let $R:SU(2)\to GL(V)$
be a complex representation and write ${W}_{m}={\oplus}_{{m}_{j}=m}{V}_{j}$
for the sum of all subspaces with weight $m$
. In other words, we're grouping the irreducible subrepresentations for the diagonal $U(1)$
into subspaces with a common weight. We'll call ${W}_{m}$
the
For ${\mathrm{Sym}}^{2}({\mathbf{C}}^{2})$ , ${W}_{2}$ , ${W}_{0}$ and ${W}_{2}$ are each 1dimensional (spanned by ${e}_{2}^{2}$ , ${e}_{1}{e}_{2}$ and ${e}_{1}^{2}$ respectively). We draw this as a weight diagram:
The following lemma tells us how $H$ , $X$ and $Y$ act in the representation:

${R}_{*}^{\mathbf{C}}(H)$ acts on ${W}_{m}$ as $mI$ , i.e. if $v\in {W}_{m}$ then ${R}_{*}^{\mathbf{C}}(H)v=mv$ .

${R}_{*}^{\mathbf{C}}(X)$ sends ${W}_{m}$ to ${W}_{m+2}$ , i.e. if $v\in {W}_{m}$ then ${R}_{*}^{\mathbf{C}}(X)v\in {W}_{m+2}$ .

${R}_{*}^{\mathbf{C}}(Y)$ sends ${W}_{m}$ to ${W}_{m2}$ .
In terms of the weight diagram earlier, we can draw arrows which show how $H$ , $X$ and $Y$ act: $H$ sends the weight spaces back to themselves, $X$ shifts to the right, and $Y$ shifts to the left. Note that in the example of ${\mathrm{Sym}}^{2}({\mathbf{C}}^{2})$ , $X$ sends ${W}_{2}$ to ${W}_{4}$ , but ${W}_{4}=0$ , so $X$ annihilates elements of ${W}_{2}$ .
This is what I meant when I said that the offdiagonal elements of $\U0001d530\U0001d532(2)$ mix up the weight spaces: they do so in a very precise way. Also, you can see why $X$ and $Y$ are more convenient than ${\sigma}_{2}$ and ${\sigma}_{3}$ : these would mix up weight spaces in a messy way that combines the $X$ and $Y$ arrows.

We have already proved part (1) of the lemma: we know that $v\in {W}_{m}$ implies ${R}_{*}({\sigma}_{1})v=imv$ , and ${\sigma}_{1}=iH$ , so ${R}_{*}^{\mathbf{C}}(H)v=mv$ .

We'll do (2) and leave the verysimilar (3) as an exercise. I'll drop the superscript $\mathbf{C}$ from now on. Suppose that $v\in {W}_{m}$ . We want to show that ${R}_{*}(X)v\in {W}_{m+2}$ . Note that $v\in {W}_{m}$ if and only if ${R}_{*}(H)v=mv$ and ${R}_{*}(X)v\in {W}_{m+2}$ if and only if ${R}_{*}(H){R}_{*}(X)v=(m+2){R}_{*}(X)v$ .
We know that ${R}_{*}(X){R}_{*}(H)v={R}_{*}(X)mv=m{R}_{*}(X)v$ , but we can't just commute ${R}_{*}(H)$ past ${R}_{*}(X)$ . However, we know that $[H,X]=2X$ , and since ${R}_{*}$ is a Lie algebra homomorphism we have $[{R}_{*}(H),{R}_{*}(X)]=2{R}_{*}(X)$ .
Therefore $${R}_{*}(H){R}_{*}(X)v={R}_{*}(X){R}_{*}(H)v+2{R}_{*}(X)v=(m+2){R}_{*}(X)v,$$ as required.
Preclass exercise
Suppose that $R:SU(2)\to GL(V)$ is a representation with weight decomposition $V=\oplus {W}_{m}$ . Show that ${R}_{*}^{\mathbf{C}}(Y)$ sends ${W}_{m}$ to ${W}_{m2}$ .