Suppose $(M, \left\langle \cdot, \cdot \right\rangle_M), (\tilde M, \left\langle \cdot, \cdot \right\rangle_{\tilde M})$ are Riemannian manifolds and $f: M \rightarrow \tilde M$ is an isometric immersion. I am trying to work with the rigorous definition of the pullback connection on $f^* T \tilde M$ to prove that the pullback connection on $f^* T \tilde M$ is the same as the Levi-Civita connection on $TM$ when restricted to the tangent part. I recall the definition I work with. If $\tilde \nabla$ is the LC connection on $T \tilde M$, we define the pullback connection on $f^* T \tilde M$ in the following way: if $\{ e_i \}$ is a local frame on $T \tilde M$, then $\{ f^* e_i \}$ is a local frame on $f^* T \tilde M$, so for a local section of $f^* T \tilde M$ written as $\sum_i a_i f^* e_i$ with $a_i$ local functions on $M$ and $X \in \Gamma(TM)$ we put: $$ \hat\nabla_{X} (\sum_i a_i f^* e_i):= \sum_i \left( X(a_i) f^*e_i + a_i f^* (\tilde\nabla_{f_*X} e_i ) \right). $$ $\hat \nabla$ is the induced connection in the pullback bundle. Let's denote the tangent part of a section $s$ in $f^* T \tilde M$ by $s^T$. Using $\hat \nabla$, we define a connection $\nabla'$ on $TM$. We note that for $Y \in \Gamma(TM)$, we can see $f_* Y$ as a section in $f^* T \tilde M$ even though it is not a section in $T \tilde M$; sections of the type $f_* Y$ span the tangent part of $f^* T \tilde M$. $$ \nabla'_X Y:= f_*^{-1} \left( \left( \hat\nabla_{f_*X} f_* Y \right)^T \right) $$
Now, I've seen here why this connection is torsion-free (this doesn't require that $f$ be an isometry and in fact even if $f$ is not an immerison, what is actually shown is that the pullback connection is always symmetric if the initial connection is symmetric).
I'm trying to also show $\nabla'$ is compatible with the Riemannian metric.
What I've tried is to argue like this. Since $f$ is an isometry, we have: $$ \left\langle \nabla'_X Y, Z \right\rangle_{M, x} = \left\langle f_* f_*^{-1} (\hat\nabla_X (f_*Y))^T, f_* Z \right\rangle_{\tilde M, f(x)} = \left\langle (\hat\nabla_X (f_*Y))^T, f_* Z \right\rangle_{\tilde M, f(x)} $$ (this equality and its derivation make sense because the Riemannian metrics are tensors, so they only require pointwise values). Similarly $$ \left\langle Y, \nabla'_X Z \right\rangle_{M, x} = \left\langle f_* Y, (\hat\nabla_X (f_*Z))^T\right\rangle_{\tilde M, f(x)} $$
For some local frame $\{ e_i \}$ of $T \tilde M$, write $f_*Y = \sum_i a_i f^*e_i$ and $f_* Z = \sum_i b_i f^* e_i$. So \begin{align*} &\left\langle (\hat\nabla_X (f_*Y))^T, f_* Z \right\rangle_{\tilde M, f(x)} + \left\langle f_* Y, (\hat\nabla_X (f_*Z))^T\right\rangle_{\tilde M, f(x)} \\ &= \left\langle \left( \sum_i \left( d_x a_i (X) e_i(x) + a_i(x) (\tilde\nabla_{f_* X} e_i)_{f(x)} \right) \right)^T , \sum_i b_i(x) e_i(f(x)) \right\rangle \\ &+ \left\langle \sum_i a_i(x) e_i(f(x)), \left( \sum_i \left( d_x b_i (X) e_i(x) + b_i(x) (\tilde\nabla_{f_* X} e_i)_{f(x)} \right) \right)^T \right\rangle \end{align*} Here I am stuck, because I have no clue how to evaluate the tangent parts which appear in the above right hand side.
I suspect this tactic is too naive. Anyway, how would one go about showing that $\nabla'$ is indeed a compatible connection?