$\newcommand{\span}{\text{span}}$ The matrix in question is: $$A=\begin{pmatrix} 2&0&0&0\\ a&-1&0&0\\ b&c&-1&0\\ d&e&f&2 \end{pmatrix}$$
This was on a test that I had and I got it wrong because I needed to add more conditions, what I did was:
Since A is a upper triangular matrix the eigenvalues of A are the values on the diagonal so $$\lambda_1=\lambda_4=2 \qquad \lambda_2=\lambda_3=-1$$ For $$\lambda_1=\lambda_4=2$$ we have:
$(A-2I)=\begin{pmatrix} 2&0&0&0\\ a&-1&0&0\\ b&c&-1&0\\ d&e&f&2 \end{pmatrix}-\begin{pmatrix} 2&0&0&0\\ 0&2&0&0\\ 0&0&2&0\\ 0&0&0&2 \end{pmatrix}= \begin{pmatrix} 0&0&0&0\\ a&-1&0&0\\ b&c&-1&0\\ d&e&f&0 \end{pmatrix}$
Since we need that the geometric multiplicity equals the algebraic multiplicity (which is $2$) we need to have 2 linearly independent vectors on each solution for the eigenvalues, so by making $a=b=c=d=e=f=0$ we get: $$\begin{pmatrix} 0&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&0 \end{pmatrix}$$
The will give us the eigenspace $E_{2}=\span\left(\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\right)$
By a similar process for $\lambda_2=\lambda_3=-1$ we get $E_{-1}=\span\left(\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\right)$
What are the remaining conditions, or values, that I need on $a,b,c,d,e,f$ for $A$ to be diagonalizable. Any help would be really appreciated.
The geometric multiplicity of the eigenvalue $\lambda$ is the dimension of the solution space of $(A - \lambda I) {\bf x} = 0$, that is, the nullity of $A - \lambda I$.
Example For our $A$, the geometric multiplicity of the eigenvalue $2$ of $A$ is the nullity of $$A - 2I = \pmatrix{0&0&0&0\\a&-3&0&0\\b&c&-3&0\\d&e&f&0} ,$$ or just as well (by the Rank-Nullity Theorem) $4$ less the rank of that matrix. So, the geometric and algebraic multiplicities of the eigenvalue $2$ are equal if and only $\operatorname{rank}(A - 2 I) = 4 - 2 = 2$. The second and third columns of $A - 2 I$ are always linearly independent (the $2 \times 2$ minor consisting of their second and third entries has determinant $9 \neq 0$), so $\operatorname{rank}(A - 2I) = 2$ if and only if the first column is a linear combination of the second and third columns, but in this case that condition is equivalent to the vanishing of the determinant of the lower-left $3 \times 3$ minor, which is a cubic condition in $a, \ldots, f$.
Repeating the process for the eigenvalue $+1$ gives another necessary set of cubic conditions, and in this case our system of conditions is equivalent to a system consisting of one quadratic condition and one linear condition in $a, \ldots, f$.