Ricci and scalar curvature

(1) implies (2) just by the definition of the trace of the linear operator $x\mapsto R(x,v)w$. In general we compute the trace of a linear operator $L$ in an inner product space $\langle \cdot, \cdot \rangle$ by taking any orthonormal basis $e_1,\ldots, e_n$ and then $\mathrm{tr}(L)=\sum_{i=1}^n \langle L(e_i), e_i\rangle$.

In your case, $\mathrm{Ric}(v,w)=\mathrm{tr}(x\mapsto R(x,v)w)=\sum_{i=1}^n g(R(e_i, v)w, e_i)$, which is (2).

The Ricci that appears in (3) is not the Ricci curvature. In fact, is the Ricci operator defined by $g(\mathrm{ric}(v),w)=\mathrm{Ric}(v,w)$. It is easy to verify that $\mathrm{ric} $ is linear so

$\mathrm{scal}=\mathrm{tr}(\mathrm{ric})=\sum_{j=1}^n g(\mathrm{ric}(e_j),e_j)=\sum_{j=1}^n \mathrm{Ric}(e_j,e_j)$ which is (because of (2)) equal to $\sum_{i,j=1}^n g(R(e_i,e_j)e_j,e_i)$.

Finally, verifying that the Sectional curvature is equal to (4) is just the definition I think.

A good reference to tensors if I'm not wrong is O Neill's book, "Semi-riemannian Geometry".

Another interesting Riemannian Geometry books apart from Petersen's book are Lee's book "Riemannian Manifolds", and Do Carmo's book "Riemannian geometry".