Relating the answer to the simpler problem
Let $I$ denote the identity matrix. Taking $v$ to be a column-vector, your first matrix can be written in the form
$$
A = \|v\|^4 I + 2\|v\|^2vv^T.
$$
This is a rank-1 perturbation of a scalar matrix, so its determinant, inverse, eigenvalues, and eigenvectors can all be easily found. In particular, the eigenvalues of $A$ are given by $\|v\|^4$ and $3\|v\|^4$.
The projection matrices $I^{\perp}$ and $I^\|$ are the spectral projections of the symmetric matrix $A$, i.e. the orthogonal projections on to the eigenspaces of $A$ associated with the respective eigenvalues. Your expressions for $A$ and $A^{-1}$ arise as a consequence of this fact.
Note that $I^\perp$ and $I^\|$ are also the spectral projections associated with the rank-1 matrix $A - \|v\|^4 I = 2\|v\|^2 vv^T$.
Justification of approach
The second matrix that you have written is
$$
B = (\|u\|^4 + \|v\|^4)I + 2 \|v\|^2 vv^T + 2\|u\|^2 uu^T + 2u^Tv(uv^T + v^Tu). \tag{1}
$$
To begin, we consider the matrix $M = B - (\|u\|^4 + \|v\|^4)I$,
$$
M = 2 \|v\|^2 vv^T + 2\|u\|^2 uu^T + 2u^Tv(uv^T + v^Tu).
$$
We can write $M$ in the form
$$
M = 2[vv^Tvv^T + vv^Tuu^T + uu^Tvv^T + uu^Tuu^T]\\
= 2(vv^T + uu^T)^2.
$$
Assuming that $u,v$ are linearly indepednent, both $(vv^T + uu^T)$ and hence $M$ will be rank-2 matrices, making $B$ a rank-2 perturbation of the identity matrix. The eigenvectors of $M$ will be the same as those of $B$.
For the remainder of this answer, I will assume that $u,v$ are linearly independent.
In order to find the eigenvectors of $M$, we begin by noting that for any vector $w \in \operatorname{span}(\{u,v\})^\perp$, we have $Mv = 0$. Because $M$ is symmetric, it follows that the remaining eigenvectors can be found within the invariant subspace $\operatorname{span}(\{u,v\})$. To that end, we set up the equation
$$
(M - \lambda I)(c_1 u + c_2 v) = 0 \iff\\
(M - \lambda I)\pmatrix{u & v}\pmatrix{c_1\\c_2} = 0 \iff\\
\pmatrix{u & v}^T(M - \lambda I)\pmatrix{u & v}\pmatrix{c_1\\c_2} = 0
$$
Let $Q$ denote the matrix $Q = [u \ \ v]$, and $c = (c_1,c_2)^T$. Because $u,v$ are linearly independent, $Q^TQ$ is invertible. We can write the above as
$$
Q^T(M - \lambda I)Q c = 0 \iff\\
(Q^TMQ - \lambda (Q^TQ)I)c = 0 \iff\\
(Q^TQ)^{-1}(Q^TMQ - \lambda (Q^TQ)I)c = 0 \iff\\
[(Q^TQ)^{-1}Q^TMQ - \lambda I] c = 0.
$$
Thus, the vectors $c$ corresponding to eigenvectors of $M$ will themselves be the eigenvectors of the $2 \times 2$ matrix $(Q^TQ)^{-1}Q^TMQ$. Once you have the two eigenvectors $y^{(1)},y^{(2)}$ of $M$ corresponding to the non-zero eigenvalues $\lambda_1,\lambda_2$, if we take $I^{\|}_1,I^{\|}_2$ to be the spectral projections
$$
I^{\|}_1 = \frac{y^{(1)}{y^{(1)T}}}{\|y^{(1)}\|^2}, \quad I^{\|}_2 = \frac{y^{(2)}{y^{(2)T}}}{\|y^{(2)}\|^2}
$$
and $I^\perp = I - I^\|_1 - I^\|_2$, then it follows that
$$
M = 0 I^\perp + \lambda_1 I^\|_1 + \lambda_2 I^\|_2,
$$
so that $B = (\|u\|^4 + \|v\|^4)I + M$ is given by
$$
B = (\|u\|^4 + \|v\|^4) I^\perp + (\|u\|^4 + \|v\|^4 + \lambda_1) I^\|_1 + (\|u\|^4 + \|v\|^4 + \lambda_2)I^\|_2.
$$
Step by step procedure
In light of the above, we can obtain a decomposition in terms of the "proper matrices" as follows: suppose that we are given vectors $u,v$ (associated with the matrix $B$ from Equation (1)). Denote $M = B - (\|u\|^4 + \|v\|^4)I$, where $I$ is the identity matrix. Denote $Q = [u\ \ v]$.
- Compute the $2 \times 2$ matrix $P = (Q^TQ)^{-1}Q^TMQ$
- Find eigenvalue/eigenvector pairs $\lambda_1, x^{(1)}$ and $\lambda_2, x^{(2)}$ of $P$
- Obtain associated eigenvectors of $B$, $y^{(i)} = x^{(i)}_1 u + x^{(i)}_2v$ for $i = 1,2$.
- Compute the spectral projection matrices,
$$
I^{\|}_1 = \frac{y^{(1)}{y^{(1)T}}}{\|y^{(1)}\|^2}, \quad I^{\|}_2 = \frac{y^{(2)}{y^{(2)T}}}{\|y^{(2)}\|^2}, \quad I^\perp = I - I^\|_1 - I^\|_2.
$$
- The matrix $B$ can be decomposed as follows:
$$
B = (\|u\|^4 + \|v\|^4) I^\perp + (\|u\|^4 + \|v\|^4 + \lambda_1) I^\|_1 + (\|u\|^4 + \|v\|^4 + \lambda_2)I^\|_2.
$$
Explicit expression for the eigenvalues
We find that $M = 2(QQ^T)^2$. It follows that
$$
(Q^TQ)^{-1}Q^TMQ = \\
2(Q^TQ)^{-1}Q^T(QQ^TQQ^T)Q = \\
2(Q^TQ)^{-1}(Q^TQ)(Q^TQQ^TQ) = \\
2(Q^TQ)^2.
$$
It follows that the desired eigenvalues are the squares of the eigenvalues of $Q^TQ$ multiplied by $2$ and that the eigenvectors are equal to the eigenvectors of $Q^TQ$. We find that
$$
Q^TQ =
\pmatrix{
u^Tu & u^Tv\\
v^Tu & v^Tv} = \pmatrix{\|u\|^2 & u^Tv\\
u^Tv & \|v\|^2}.
$$
With the quadratic formula, we find that the eigenvalues are given by
$$
\lambda_1 = \frac 12 \left(\|u\|^2 + \|v\|^2 + \sqrt{(\|u\|^2 + \|v\|^2)^2 - 4[\|u\|^2\|v\|^2 - (u^Tv)^2]}\right)^2\\
\lambda_2 = \frac 12 \left(\|u\|^2 + \|v\|^2 - \sqrt{(\|u\|^2 + \|v\|^2)^2 - 4[\|u\|^2\|v\|^2 - (u^Tv)^2]}\right)^2,
$$
which simplifies further to
$$
\lambda_1 = \frac 12 \left(\|u\|^2 + \|v\|^2 + \sqrt{(\|u\|^2 - \|v\|^2)^2 + 4(u^Tv)^2}\right)^2\\
\lambda_2 = \frac 12 \left(\|u\|^2 + \|v\|^2 - \sqrt{(\|u\|^2 - \|v\|^2)^2 + 4(u^Tv)^2}\right)^2.
$$
In the case where $u^Tv \neq 0$, the associated eigenvectors can be written as
$$
x^{(i)} = \pmatrix{u^Tv \\
\sqrt{\lambda_i/2} - \|u\|^2}, \qquad i =1,2.
$$
