Let $M$ be a smooth manifold with or without boundary, and let $\nabla$ be a connection in $TM$. Suppose we are given two smooth local frames $(E_i)$ and $(\widetilde{E}_j)$ for $TM$ on an open subset $U \subset M$ , related by $\widetilde{E}_i = A^j_i E_j$ for some matrix of functions $A^j_i$. Let $\Gamma^k_{ij}$ and $\widetilde{\Gamma}^k_{ij}$ denote the connection coefficients of $\nabla$ with respect to the frames. Show that $$\widetilde{\Gamma}^k_{ij} = (A^{-1})^k_pA_i^qA^r_j\Gamma^p_{qr} +(A^{-1})^k_pA_i^qE_q(A^p_j).$$
This is quite overwhelming with the indices and in Lee's book on Riemannian manifolds prior to this there isn't a definition for the connection coefficients so I think I need to start from $\nabla_XY$ where $X=X^i\widetilde{E}_i$ and $Y=Y^j\widetilde{E}_j$. What I have is that $$\nabla_XY=X(Y^j)\widetilde{E}_j + X^iY^j\widetilde{\Gamma}^k_{ij}\widetilde{E}_k.$$ Expanding the $\widetilde{E}_j$'s I have that $$\nabla_XY=X(Y^j)A^q_jE_q + X^iY^j\widetilde{\Gamma}^k_{ij}A^p_kE_p$$ and so $$X^iY^j\widetilde{\Gamma}^k_{ij}A^p_kE_p = \nabla_XY-X(Y^j)A^q_jE_q $$
but I don't know how to isolate $\widetilde{\Gamma}^k_{ij}$ from this expression?
This is standard enough. Let me write it out. To get the formula for $\tilde\Gamma_{ij}^k$, don't introduce any new vector fields. Just work with the basis vector fields $\tilde E_i$.
By definition, $$ \nabla_{\tilde E_i} \tilde E_j =\tilde \Gamma_{ij}^k \tilde E_k. $$ Since the $\tilde E_k$ are basis, the vector fields $\nabla_{\tilde E_i} \tilde E_j$ must have such unique expression in them.
We now compute, using the properties of $\nabla$: \begin{align} \nabla_{\tilde E_i} \tilde E_j &= \nabla_{A_i^q E_q} (A_j^r E_r) = A_i^q \nabla_{E_q}(A_j^r E_r)\\ &=A_i^q \big(E_q(A_j^r)E_r + A_j^r\nabla_{E_q}E_r\big)\\ &=A_i^q \big(E_q(A_j^r)E_r + A_j^r\Gamma_{qr}^p E_p\big)\\ &=A_i^q \big(E_q(A_j^r)(A^{-1})^k_r \tilde E_k + A_j^r\Gamma_{qr}^p (A^{-1})^k_p \tilde E_k\big)\\ &=\big((A^{-1})^k_p A_i^q A_j^r \Gamma_{qr}^p + (A^{-1})^k_r A_i^q E_q(A_j^r)\big)\tilde E_k\\ &=\big((A^{-1})^k_p A_i^q A_j^r \Gamma_{qr}^p + (A^{-1})^k_p A_i^q E_q(A_j^p)\big)\tilde E_k \end{align} In this last line, we just changed the index $r$ in the second term to $p$, for the indices to match better. This gives the answer, since the whole coefficient is $\tilde \Gamma_{ij}^k$.
Also note the first term is the tensorial transformation part of $\Gamma$, and the second term says how it differs from a tensor.