Computing the $2^\text{nd}$ fundamental form of $\mathbb{R}_+\times M^n\to \mathbb{R}^{n+p+1}$ 
Let $f: M^n\to \mathbb{S}^{n+p}$ be an isometric immersion. The cone over $f$ is defined to be the immersion
  \begin{align*}
F:\mathbb{R}_+\times M &\to \mathbb{R}^{n+p+1}\\
(t,x)&\mapsto tf(x)
\end{align*}
  Compute the second fundamental form of $F$.
[where $\mathbb{R}_+:=\{t\in\mathbb{R}\mid t>0\}$]

I could verify that $F$ is an immersion by considering $f(x)=(f_1(x),...,f_{n+p+1}(x))\in\mathbb{R}^{n+p+1}$ with $\sum_if_i^2=1$. That way, $\langle\frac{\partial f}{\partial x_i},f\rangle=0$ for all $i$ and, since $f$ is an immersion, we have $\frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n}$ linearly  independent, so $\text{rank}(dF)=\text{rank}\left(f,\frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n}\right)=n+1$. 
For the $2^\text{nd}$ fundamental form, here's where I'm at: If $\widetilde{\nabla}$, $\nabla$ are Levi-Civita connections for $\mathbb{R}^{n+p+1}$ and $\mathbb{R}_+\times M$ respectively, the second fundamental form is by definition $\alpha(X,Y)=\widetilde{\nabla}_\widetilde{X}\widetilde{Y}-\nabla_XY$. By tensoriality and symmetry of $\alpha$, we only need to compute $\alpha\left(\frac{\partial}{\partial t},\frac{\partial}{\partial t}\right),\alpha\left(\frac{\partial}{\partial t},\frac{\partial}{\partial x_i}\right)$ and $\alpha\left(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\right)$.
To compute $\alpha\left(\frac{\partial}{\partial t},\frac{\partial}{\partial t}\right)$ for example, can see that $\nabla_{\frac{\partial}{\partial t}}\frac{\partial}{\partial t}=0$, but I don't know how compute $\widetilde{\nabla}_{\frac{\partial F}{\partial t}}\frac{\partial F}{\partial t}$. Obviously, $\frac{\partial F}{\partial t}(t,x)=f(x)$, but I can't figure out what $\widetilde{\nabla}_{f(x)}f(x)$ even means.
Any suggestions?
 A: The question is purely local, so we may choose a coordinate chart $\varphi \colon U \subset \mathbb{R}^n \to M$ around some point $x \in M$. We use this chart to indentify points in $U$ and $\varphi(U)$, and thus we regard $f$ locally as $\varphi \circ f$, so that the mapping $f \colon U \to \mathbb{R}^{n+p+1}$ is a parametrization of its image, and
$$
\langle f, f \rangle = 1
$$
The mapping
$F \colon \mathbb{R}_{+} \times U \to \mathbb{R}^{n+p+1} \colon (t, x) \mapsto t f(x)$
is also a local parametrization of its image, which is a piece of the cone $C = F(\mathbb{R}_{+} \times U)$.
We calculate at the point $(t, x) \in C$.
Here is a picture showing schematically what is going on:

Corresponding to the chart $\varphi$, there are local coordinates $x^i$ in $U$, and let $t$ be a local coordinate in $\mathbb{R}_{+}$, so that $\{ t, x^i, i=1, \dots, n \}$ form a coordinate system on $\mathbb{R}_{+} \times U$.
Let us introduce the following notation: $\partial_i := \frac{\partial}{\partial x^i}$, $\partial_t := \frac{\partial}{\partial x^i}$, $f_i := \partial_i f$, $F_i := \partial_i F$, $F_t := \partial_t F$, $i = 1, \dots, n$. Since $F = t f$, we have:
$$
F_i = tf_i \text{ and }F_t = f
$$
Notice that $F_i$ and $F_t$ are vector fields along the cone $C$, and
$$
F_i|_{(t,x)} = t \pmatrix{
f^1_i(x)\\
\vdots\\
f^{n+p+1}_i(x)
}
\text{ and }
F_t|_{(t,x)} = \pmatrix{
f^1(x)\\
\vdots\\
f^{n+p+1}(x)
}
$$
In order to differentiate $F_i$ and $F_t$ in $\mathbb{R}^{n+p+1}$ we need some local coordinates there and some extensions $\widetilde{F}_i$ and $\widetilde{F}_t$ of those fields, so we just take the slice coordinates in $\mathbb{R}_{+} \times U \times \mathbb{R}^{p}$
$$
y^0 = t, y^1 = x^1, \dots, y^n = x^n, y^{n+1} = y^{n+1}, \dots, y^{n+p} = y^{n+p}
$$
(where $y^{n+1}, \dots, y^{n+p}$ are the standard coordinates in $\mathbb{R}^{p}$), and define
$$
\widetilde{F}(t, x^1, \dots, x^n, *, \dots, *) := F(t, x^1, \dots, x^n)
$$
so that $\widetilde{F}_i = \partial_i \widetilde{F}$ and $\widetilde{F}_t = \partial_i \widetilde{F}$.
Since in $\mathbb{R}^{n+p+1}$ we have the standard Euclidean metric and the standard Euclidean connection (all Christoffel symbols vanish identically),
we have
$$
\widetilde{\nabla}_i \widetilde{F}_j = \partial_i F_j \\
\widetilde{\nabla}_t \widetilde{F}_i = \partial_t F_i \\
\widetilde{\nabla}_t \widetilde{F}_t = \partial_t F_t
$$
In order to compute the s.f.f. $\alpha(X,Y)=\widetilde{\nabla}_\widetilde{X}\widetilde{Y}-\nabla_XY$ it remains to understand what $\nabla_XY$ is here.
It is known that $\nabla_XY = (\widetilde{\nabla}_\widetilde{X}\widetilde{Y})^{\top} = \Pi(\widetilde{\nabla}_\widetilde{X}\widetilde{Y})$, where the tangential projection operator is given by
$$
\Pi \colon v \mapsto \frac{1}{|F_t|^2} \langle v, F_t \rangle F_t + \sum_{i=1}^{n} \frac{1}{|F_i|^2} \langle v, F_i \rangle F_i
$$
This is almost it, because now we can write
$$
\nabla_i F_j = \Pi(\widetilde{\nabla}_i \widetilde{F}_j) = \frac{1}{|F_t|^2} \langle \widetilde{\nabla}_i \widetilde{F}_j, F_t \rangle F_t + \sum_{k=1}^{n} \frac{1}{|F_k|^2} \langle \widetilde{\nabla}_i \widetilde{F}_j, F_k \rangle F_k \\
\nabla_t F_i = \Pi(\widetilde{\nabla}_t \widetilde{F}_i) = \frac{1}{|F_t|^2} \langle \widetilde{\nabla}_t \widetilde{F}_i, F_t \rangle F_t + \sum_{k=1}^{n} \frac{1}{|F_k|^2} \langle \widetilde{\nabla}_t \widetilde{F}_i, F_k \rangle F_k \\
\nabla_t F_t = \Pi(\widetilde{\nabla}_t \widetilde{F}_t) = \frac{1}{|F_t|^2} \langle \widetilde{\nabla}_t \widetilde{F}_t, F_t \rangle F_t + \sum_{k=1}^{n} \frac{1}{|F_k|^2} \langle \widetilde{\nabla}_t \widetilde{F}_t, F_k \rangle F_k
$$
and collecting all the pieces it is now straightforward to obtain the final expressions.
