do Carmo Riemannian Geometry Exercise 2.3: definition of $\nabla$ for an immersion - Part II The following is Exercise 3 of Chapter 2 of my Brazilian edition of do Carmo's Riemannian Geometry:

Let $f: M^n \to \overline M^{n + k}$ be an immersion from a differentiable manifold $M$ to a Riemannian manifold $\overline M$. Assume in $M$ the Riemannian metric induced by $f$:
$$
\langle u, v \rangle_p = \langle df_p(u), df_p(v) \rangle_{f(p)}.
$$
Let $p \in M$ and $U \subset M$ be a neighborhood of $p$ such that $f(U) \subset \overline M$ be a submanifold of $\overline M$. Let $X, Y$ be vector fields on $f(U)$ and extend then to vector fields $\overline X, \overline Y$ on an open subset of $\overline M$. Define
$$
(\nabla_X Y)(p) = \text{tangential component of } \overline \nabla_{\overline X} \overline Y(p), 
$$
where $\overline \nabla$ is the Riemannian connection of $\overline M$. Prove that $\nabla$ is the Riemannain connection of $M$.

As already discussed in this question, the notation is very confusing and in fact wrong. Proceed as in the answer to the linked question:
if $X, Y$ are vector fields on $U$, we consider the vector fields $df(X), df(Y)$ on $f(U)$ and extend these vector fields to $\overline X, \overline Y$ on an open set of $\overline M$. Recall also that we have the following decomposition of the tangent space at $f(p)$:
$$
T_{f(p)}\overline M = df_p(T_pM) \oplus (df_p(T_pM))^\perp,
$$
and we call the tangential component of $\overline \nabla_{\overline X} \overline Y$ the terms in $df_p(T_pM)$. So the correct definition for $\nabla$ is
$$
\nabla_X Y(p) = (df_p)^{-1}( \text{tangential component of }\overline \nabla_{\overline X} \overline Y(f(p))).
$$
First we show that $\nabla$ well-defined, i.e., does not depend on the extensions $\overline X, \overline Y$. Indeed, if $\overline X_1, \overline X_2, \overline Y_1, \overline Y_2$ are two distinct extensions of $df(X), df(Y)$ respectively, then they coincide at $f(p)$. Then their tangential components coincide and $\nabla$ is indeed well-defined.
Now, in order to show that $\nabla$ is a connection, we have to show the three defining properties. Don't we need that $f$ be in fact an embedding, in order to be able to define $g \circ f^{-1}$ to prove, for example, $\nabla_{g X + h Y}Z = g \nabla_X Z + h \nabla_Y Z$?
Also, how to show that this connection is compatible with the Riemannian metric?
Finally, is the following argument for symmetry correct?
Ordering the basis in $T_{f(p)}\overline M$ so that $X_1, \ldots, X_n \in df_p(T_pM)$:
\begin{align*}
\nabla_{X_i} X_j - \nabla_{X_j} X_i & = (df_p)^{-1}\left(\sum_k\Gamma_{ij}^k X_k\right) - (df_p)^{-1}\left(\sum_k\Gamma_{ji}^k X_k\right) \\
& = (df_p)^{-1} \left(\sum_k(\Gamma_{ij}^k-\Gamma_{ji}^k) X_k \right) \\ 
& = 0
\end{align*}
by the symmetry of $\overline \nabla$. Thus $\nabla$ is symmetric.
 A: First, your checking that $\nabla_XY$ is well defined independent of $\overline X, \overline Y$ is unclear: to take an analogy, even if two functions $f_1, f_2$ agree at a point $p$, it does not imply that $f'_1 = f_2'$ at $p$.
To check that $\nabla$ is well-defined, we split into two steps:

*

*If $\overline X, \widetilde X$ are both extension of $df(X)$, then for any local vector fields $Z$ on $V\subset \overline M$ and for all $p\in U$,
$$ \overline \nabla_{\overline X} Z = \overline \nabla_{\widetilde X} Z\ \ \ \ \ \ \text{ at } f(p).$$
Proof: This follows from the fact that $\overline \nabla$ is $C^\infty$-linear in that component, thus the value $\overline \nabla_{\overline X} Z(f(p))$ depends only on $\overline X(f(p))$.


*Let $\overline Y, \widetilde Y$ are both extension of $df(Y)$ and $\overline X$ is tangential to $f(U)$, then
$$\tag{2} \overline \nabla _{\overline X} \overline Y =  \overline \nabla _{\overline X} \widetilde Y\ \ \ \ \ \text{ at }f(p).$$
Proof: this follows from the fact that covariant differentiation can be computed using parallel transport (here): In particular, since $\overline X$ is tangential to $f(U)$, one can find an integral curve of $\overline X$ which lies inside $f(U)$ (For example, let $\gamma : (-\epsilon, \epsilon)\to M$ be an integral curve of $X$. Then $f\circ \gamma$ is an integral curve of $\overline X$ lying inside $f(U)$). Since $\overline Y, \widetilde Y$ agrees on $f(U)$, (2) is shown.
Second, we show that $\nabla$ is indeed a connection. To begin with, we show
(1) For any local vector fields $X, Y$ on $U$ and local smooth functions $\varphi:U \to \mathbb R$, we have
$$\nabla_{\varphi X} Y (p) = \varphi(p) \nabla_X Y(p), \ \ \ \forall p\in U.$$
Proof: let $\overline \varphi$ be a smooth function on $V\subset \overline M$ which extends $\varphi\circ f^{-1} : f(U) \to \mathbb R$. That is, for all $f(p) \in f(U)$ we have
$$ \varphi (p) = \overline \varphi (f(p)).$$
Then $\overline \varphi \overline X$ is an extension of $df (\varphi X)$. So
\begin{align*}
\nabla _{\varphi X} Y(p) &= df^{-1} \bigg(\text{tangential component of } \overline \nabla_{\overline\varphi \overline X} \overline Y(f(p))\bigg) \\
&= df^{-1} \bigg(\text{tangential component of }\ \overline\varphi (f(p)) \overline \nabla_{\overline X} \overline Y(f(p))\bigg) \\
&= \varphi (p) df^{-1} \bigg(\text{tangential component of }\overline \nabla_{\overline X} \overline Y(f(p))\bigg) \\
&= \varphi (p) \nabla_X Y (p). 
\end{align*}
(2) We show also that $\nabla$ is compatible with the pullback metric $g = f^*\bar g$, let $X, Y, Z$ be vector fields. Then by definition,
$$ X g(Y, Z)(p) = \frac{d}{dt}\bigg|_{t=0} g_{\gamma(t)}( Y(\gamma(t)), Z(\gamma(t))),$$
where $\gamma : (-\epsilon, \epsilon) \to M$ is any curve with $\gamma(0) = p$, $\gamma'(0) = X(p)$. Using the definition of pullback metric,
$$ g_{\gamma(t)}( Y(\gamma(t)), Z(\gamma(t))) = \bar g_{f(\gamma(t))} (df_{\gamma(t)} Y(\gamma(t)), df_{\gamma(t)} Z(\gamma(t))).$$
Since $f\circ \gamma$ is a curve in $\overline M$ with $f\circ \gamma (0) = f(p)$, $(f\circ \gamma)'(t) = df_{\gamma(t)} X(\gamma(t))$, we have
\begin{align*}
\frac{d}{dt}\bigg|_{t=0} g_{\gamma(t)}( Y(\gamma(t)), Z(\gamma(t)))&= \overline X \bar g (\overline Y, \overline Z) f(p)\\
&= \bar g(\overline \nabla _{\overline X} \overline Y , \overline Z ) + \bar g(\overline Y , \overline \nabla _{\overline Y} \overline Z) \ \ \ \ \ \text{ at } f(p)\\
&= \bar g(df (\nabla _{X} Y) , df ( Z) ) + \bar g(df(Y) , df(\nabla _{Y} Z) \\
&= g(\nabla_X Y, Z) + g(Y, \nabla_XZ) 
\end{align*}
at $p$. Note we used that $\overline Y, \overline Z$ are tangential to $f(U)$, so that we have
$$ \bar g (\overline \nabla_{\overline X} \overline Y, \overline Z) = \bar g ((\overline \nabla_{\overline X} \overline Y)^\top, \overline Z),$$
where $(\cdot)^\top$ denotes the tangential part of a vector.
Finally, in your checking of the symmetry of $\nabla$ you used $\Gamma_{ij}^k = \Gamma_{ji}^k$, which a priori you do not know yet. Indeed the symmetry of $\nabla$ is equivalent to the symmetry of $\Gamma$.
To give a correct proof we, just like all the others properties we proved, pushforward everything to $\overline M$, prove the property there and then pullback: by definition,
\begin{align*}
\nabla_X Y- \nabla_Y X &= df^{-1} \left( \overline\nabla_{\overline X} \overline Y - \overline\nabla _{\overline Y} \overline X\right)^\top \\
&=  df^{-1} ([\overline X, \overline Y]^\top).
\end{align*}
Since $f(U)$ is a submanifold and $\overline X, \overline Y$ are tangential to $f(U)$,
$$ [\overline X, \overline Y]^\top = [\overline X, \overline Y] = [df (X), df(Y)]$$
(this can be check directly, assuming that $f(U)$ is a plane $\mathbb R^n \subset \mathbb R^{n+k}$. The Riemannian structure is not used here). Then by this, we have
$$\nabla_X Y- \nabla_Y X = [X, Y].$$
