In order to determine how measuring one value for $U_1$ (say $\lambda_{1,1}$) affects the probability of measuring a particular value for $U_2$ I would want to write the particle in terms of a new basis given by the eigenvectors to get the probability of each outcome, right?
Pretty much, yes.
Given two observables (Hermitian operators) $\newcommand{\calO}{{\mathcal O}}\calO_1,\calO_2$, "measuring $\calO_1$" results in collapsing the state you are measuring, call it $|\psi\rangle$, in one of the eigenvectors of $\calO_1$, call this $|m_1\rangle$.
If you then immediately afterwards measure $\calO_2$, to find the probabilities of finding each one of the eigenvectors of $\calO_2$ you compute the squared overlap between $|m_1\rangle$ and said eigenvectors.
About the general question of how measurements affect each other, it's hard to answer in full generality because it will depend a lot on the context and on what exactly you mean with "how".
For example, if you are measuring a single-qubit state and the observables considered are two Pauli matrices, say $\calO_1=X$ and $\calO_2=Y$ (observe how they do not commute), then after the first measurement you will have maximal uncertainty about the results of the second measurement (in other words, the outcome probabilities for the second measurement are always $50/50$).
This happens because any pair of eigenvectors of $X$ and $Y$ have the same squared overlap (we say that the corresponding bases are mutually unbiased).
The canonical answer about why the commutator between the observables is relevant for their being compatible is the uncertainty principle. One can show that, given any pair of observables $A$ and $B$, you have
$$\sigma_A \sigma_B \ge \lvert\langle AB\rangle - \langle A\rangle\langle B\rangle\rvert,$$
where $\sigma_A^2\equiv \langle A^2\rangle-\langle A\rangle^2$ are the variances. The right-hand side is lower bounded by the (expectation value of) the commutator $[A,B]$, which is the standard form of Heisenberg's uncertainty principle.
