What does this notation for spin mean? $\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}=\mathbf{1}\oplus\mathbf 0$

This is a really deep question and I urge you to go ahed and read about it in the literature i'll give at the end. I'll try to give a glimpse of what this actually means.

In physics we can construct our theories based solely upon symmetries of a system. When talking about angular momentum and spin in non relativistic quantum mechanics, we are talking about a specific set of symmetry, namely $SU(2)$ symmetry. $SU(2)$ is a lie group and, being a group, it's an abstract object. To make it something useful we use, what is called as a representation. There are many representation of $SU(2)$ and the one we're interested in is the spinorial representation.

The spinorial is the fundamental representation of $SU(2)$ since all representations can be constructed from tensor product of spinors. In physical terms this means that you can construct composite systems just by using spin $1/2$ particles. What you gave is how to construct a spin $1$ or spin $0$ from two spin $1/2$ systems $$\mathbf{\frac{1}{2}\otimes\frac{1}{2} = 0\oplus 1} $$

what do this numbers indicate? A number written in boldface gives the dimension (which is $2j+1$ where $j$ is the boldface number) of an irriducible representation of that group. What this implies is that you can decompose a composite system of two spin $1/2$ particles into two irriducible representation of a spin $0$ system and a spin $1$ system.

If all of this seems confusing, it's normal, it's a lot of stuff. I would suggest the following reading if you want to get a better undestanding

• Lie algebras in particle physics, Georgi
• Group Theory in a Nutshell for Physicists, Zee
• Group Theory, a physicist's survey, Ramond

The $\otimes$ sign denotes the tensor product. Given two matrices (let’s say $2\times 2$ although they can be $n\times n$ and $m\times m$) $A$ and $B$, then $A\otimes B$ is the $4\times 4$ matrix \begin{align} A\otimes B =\left( \begin{array}{cc} A_{11}B&A_{12}B\\ A_{21}B&A_{22}B \end{array}\right)= \left(\begin{array}{cccc} A_{11}B_{11}&A_{11}B_{12}&A_{12}B_{11}&A_{12}B_{12}\\ A_{11}B_{21}&A_{11}B_{22}&B_{12}B_{21}&A_{12}B_{22}\\ A_{21}B_{11}&A_{21}B_{12}&A_{22}B_{11}&A_{22}B_{12}\\ A_{21}B_{21}&A_{21}B_{22}&A_{22}B_{21}&A_{22}B_{22} \end{array}\right) \, . \end{align} A basis for this space is spanned by the vectors \begin{align} a_{1}b_{1}&\to \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)\, ,\quad a_1b_2 \to \left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right)\, ,\quad a_2b_1\to \left(\begin{array}{c} 0 \\ 0 \\ 1 \\0\end{array}\right)\, ,\quad a_2b_2\to \left(\begin{array}{c} 0\\0\\0\\1\end{array}\right) \end{align} In terms of $a_1\to \vert +\rangle_1$, $a_2\to \vert -\rangle_1$ etc we have \begin{align} a_1b_1\to \vert{+}\rangle_1\vert {+}\rangle _2\, ,\quad a_1b_2\to \vert{+}\rangle_1\vert{-}\rangle _2 \, ,\quad a_2 b_1\to \vert{-}\rangle_1\vert {+}\rangle _2 \, ,\quad a_2b_2\to \vert{-}\rangle_1\vert{-}\rangle_2\, . \end{align} In the case of two spin-$1/2$ systems, $\frac{1}{2}\otimes \frac{1}{2}$ implies your are taking $\sigma_x\otimes \sigma_x$, $\sigma_y\otimes \sigma_y$, $\sigma_z\otimes \sigma_z$, since these are operators acting on individual spin-$1/2$ systems. The resulting matrices can be simultaneously block diagonalized by using the basis states $a_1b_1$, $\frac{1}{\sqrt{2}}(a_1b_2\pm a_2b_1)$ and $a_2b_2$. There is a $3\times 3$ block consisting of $a_1b_1, \frac{1}{\sqrt{2}}(a_1b_2+a_2b_1)$ and $a_2b_2$ and a $1\times 1$ block with basis vector $\frac{1}{\sqrt{2}}(a_1b_2-a_2b_1)$.

The $3\times 3$ block never mixes with the $1\times 1$ block when considering the operators $S_x=s_x^{1}+s_x^{2}$ etc. The basis vectors of the $3\times 3$ block transform as states with $S=1$, in the sense that matrix elements of $S_x$, $S_y$ and $S_z$ are precisely those of states with $S=1$; the basis vector of the $1\times 1$ block transforms like a state of $S=0$. Hence one commonly writes \begin{align} \frac{1}{2}\otimes \frac{1}{2} = 1\oplus 0 \end{align} with the $\oplus$ symbol signifying that the total Hilbert space is spanned by those vectors spanning the $S=1$ block plus the vector spanning the $S=0$ part; note that those vectors are product states of the type $a_1b_1$ etc.

I should concur with the other answers that there is no substitue for reading up in good texts and WP.

You are right that, in a given basis, there is a similarity (equivalence) transformation implied in the equation of your title: it basically means that the tensor product on the l.h.s. is reducible, by an orthogonal basis change to the r.h.s.; that is, in words,

• The Kronecker product of two 2-vectors (spinors; in general you have (2s+1)-dim vectors!) is a 4-vector. But rotations keep two subspaces in it separate: a 3-vector subspace, and a 1-vector (scalar) subspace. However, this is invisible to the naked eye. There is an orthogonal basis change, the Clebsch matrix, which visibly separates these two subspaces, so rotations act on these visibly separately, by block matrix action. (In your singlet case, by no action at all! the rotation matrices are the identity, 1).

Can you find this 4×4 Clebsch matrix $\cal C$ in Problem 4 here for your exact problem? (Hint: mix up just the 2nd and 3rd components by a rotation of $\pi/4$.) The "right is detail in left" convention in the tensor product amounts to $$ \begin{pmatrix} a_1\\a_2\end{pmatrix} \otimes \begin{pmatrix} b_1\\b_2\end{pmatrix} = \begin{pmatrix} a_1 b_1\\a_1 b_2 \\ a_2 b_1\\ a_2 b_2\end{pmatrix} \leadsto \begin{pmatrix} \uparrow \uparrow\\ \uparrow \downarrow \\ \downarrow \uparrow\\ \downarrow \downarrow \end{pmatrix} , $$ in the spherical basis notation. The second and 3rd component, then mix up into $(\frac{\uparrow \downarrow + \downarrow\uparrow }{\sqrt{2}},\frac{\uparrow \downarrow - \downarrow\uparrow }{\sqrt{2}} )$, the triplet component and the singlet component.

The upshot is a direct sum of a 3-vector (components 1,2, & 4) and a singlet (component 3): $$ \begin{pmatrix} \uparrow \uparrow\\ \frac{\uparrow \downarrow + \downarrow\uparrow }{\sqrt{2} } \\ \downarrow \downarrow \end{pmatrix} \oplus \frac{\uparrow \downarrow - \downarrow\uparrow }{\sqrt{2}} . $$

Your title formula, however, never picks a basis.

Finally, there are elaborate formulas for recursive compositions of spins, pioneered by Bethe and elaborated by several authors afterwards. Your case is particularly simple, as WP details. I copy the WP formula, which uses dimensionality, instead of spin indices (2s+1 instead of your s), since you can do instant arithmetic checks by ignoring the circles in × and + !

Combining n doublets (your spin 1/2s) nets you $$ {\mathbf 2}^{\otimes n} = \bigoplus_{k=0}^{\lfloor n/2 \rfloor}~ \Bigl( {n+1-2k \over n+1} {n+1 \choose k}\Bigr)~~({\mathbf n}+{\mathbf 1}-{\mathbf 2}{\mathbf k})~,$$ where $\lfloor n/2 \rfloor$ is the integer floor function; the number preceding the boldface irreducible representation dimensionality (2 s+1) label indicates multiplicity of that representation in the representation reduction. The random walk that takes you there reconstructs the celebrated Catalan's triangle.

For instance, from this formula, addition of three spin 1/2 s yields a spin 3/2 and two spin 1/2s, ${\mathbf 2}\otimes{\mathbf 2}\otimes{\mathbf 2}={\mathbf 4} \oplus{\mathbf 2}\oplus{\mathbf 2} $; four spin 1/2 s yields two singlets, three spin 1 s, and one spin 2, and so forth.