I’ll just write up an answer with a few ways of attacking the problem, and let you fill in some of the details (which can easily be found if you search the site or any decent book). First, let me make the remark that we rarely use definitions to actually compute things. We use definitions to prove theorems (and then we use those theorems and definitions to prove more theorems and make up new definitions… and the cycle continues the deeper down the rabbit hole you go). To actually compute things, we invoke theorems (and some of the most important calculations are results of very important theorems; carefully justifying certain calculations can lead to a whole branch of math). Having that preliminary philosophy out of the way, let’s get to the solution.
Smoothness
You may wonder why the map in question actually smooth. Well, the answer to this is again by a theorem (whose proof is essentially by unwinding definitions, and can be found in many differential geometry books):
Theorem 0. (Smoothness in manifold vs vector space setting)
Let $V,W$ be (say finite-dimensional for simplcity) real normed vector spaces, and $A\subset V$ an open set, and $f:A\to W$ a given map. Then, $f$ is smooth in the usual multivariable calculus sense (see this for a whole bunch of equivalences) if and only if $f$ is smooth in the manifold sense (when $V,W$ are given the usual smooth structures consisting of the identity chart (or depending on your choice of definitions, a chart consisting of a linear isomorphism $V\to\Bbb{R}^{\dim V}$, and likewise for $W$)).
Theorem 1. (Smoothness of restrictions)
Let $M_1,M_2$ be smooth manifolds, $S_1\subset M_1,S_2\subset M_2$ smooth embedded submanifolds, and $f:M_1\to M_2$ a map such that $S_1$ is mapped into $S_2$. Let $\tilde{f}:S_1\to S_2$ denote the induced restriction of $f$. If $f$ is smooth, then so is $\tilde{f}$.
In your case, take $M_1, M_2$ to be respectively, the space of $n\times n$
and $(n+1)\times (n+1)$ matrices of real numbers, and $f(A)=\begin{pmatrix}\det A&0\\0&A\end{pmatrix}$. This is smooth because $\det$ is smooth (it is even an analytic function, since it is even a polynomial in the matrix entries). So, the induced map $\text{O}(n)\to \text{SO}(n+1)$ is also smooth.
Fastest Solution:
With the above remarks in mind, here’s the fastest solution to the problem
A smooth Lie group homomorphism has constant rank, so if it is injective then by the global rank theorem, it is an immersion.
These facts can definitely be found in Lee’s book. This is a very nice theorem and is one example of the many ‘automatic’ results for Lie groups. This seems like magic, but the fact that Lie group homomorphisms have constant rank is pretty easy (keep the $f(ax)=f(a)f(x)$ example from the comments in mind for intuition); the global rank theorem does the heavy lifting for us (you need to get a good grasp on the inverse and implicit function theorems, and constant rank theorem (and their equivalences), and their consequences, such as the global rank theorem).
A more naive solution.
Here, let us keep in mind that Lie group homomorphisms have constant rank; this is easy to prove. The tough part above was the use of the global rank theorem, so let us avoid this here. So, now, we have to actually compute the tangent map at the identity. Well, before this, you have to understand how the tangent spaces of submanifolds are related to tangent spaces of the ambient manifold, and the very important special case when the ambient manifold is actually a vector space. I’ve written several answers on this matter, so I suggest you look through them: How is the directional derivative used to determine the tangent map?, Calculating the derivative of a differentiable map between manifolds, Is tangent vector on so(3) a matrix?, and the various sublinks.
With all that in mind, you should now know that the Lie algebra $\mathfrak{o}(n)$ consists of (or more precisely, depending on how you define things, can very easily be identified with, using an appropriate isomorphism) the $n\times n$ real skew-symmetric matrices. Also, you should now know how the tangent map between the tangent spaces of the submanifolds is related to the usual Frechet derivative of the ambient map (you restrict the Frechet derivative of the ambient map, and thena. certain diagram commutes, as I describe in the links).
So, now, let us compute the derivative of the original map between vector spaces. We have
\begin{align}
f(A)&:=\begin{pmatrix}
\det A&0\\
0& A
\end{pmatrix},
\end{align}
so for any $\xi\in M_{n\times n}(\Bbb{R})$, we have
\begin{align}
Df_I(\xi)&=
\begin{pmatrix}
D(\det)_I(\xi)&0\\
0& D(\text{id}_{M_{n\times n}(\Bbb{R})})_I(\xi)
\end{pmatrix}
=
\begin{pmatrix}
\text{tr}(\xi)&0\\
0&\xi
\end{pmatrix}.
\end{align}
The reason why the derivative of determinant is the trace can be found on e.g in the second half of my answer here, or simply by googling it (there are many ways to prove it). Also, the identity map is a linear transformation, so at every point it is its own derivative.
So, $Df_I:M_{n\times n}(\Bbb{R})\to M_{(n+1)\times (n+1)}(R)$ is also clearly injective. Hence, its restriction to any subspace is also an injective linear map. So, here we are fortunate enough that we didn’t even need to know what explicitly the Lie algebra of the orthogonal group is; our calculation (and remarks above) shows that the tangent map between the Lie algebras is injective.
For completeness, I’ll mention that if you restrict $\xi$ to be in $\mathfrak{o}(n)$, then it is skew-symmetric, so all its diagonal entries are $0$, it has zero trace, and so $Df_I(\xi)=\begin{pmatrix}0&0\\0&\xi\end{pmatrix}$, which up to composing with isomorphisms, is precisely the tangent map between the Lie algebras (and as a sanity check, you can see that this resulting matrix is also skew-symmetric, as it should, since it must lie in the Lie algebra of $\text{SO}(n+1)$).
Finally, let’s say you don’t even want to invoke the easy fact that Lie group homomorphisms have constant rank. Then you’d have to compute $Df_A$ for an arbitrary $A\in \text{O}(n)$, and show that its restriction to the tangent space of the group at $A$ (which by the linked answers, can be identified with an honest subspace of your ambient vector space) is injective. I’ll just tell you the answer for the derivative (since the details can be found easily online) that in general, for an invertible $A\in GL(n)$, and any $\xi\in M_{n\times n}(\Bbb{R})$ we have
\begin{align}
Df_A(\xi)&=
\begin{pmatrix}
(\det A)\cdot \text{tr}(A^{-1}\xi)&0\\0&\xi
\end{pmatrix}.
\end{align}
This is once again clearly injective, so its restriction to any subspace is also injective. So, here we’re fortunate that we don’t need to know what exactly the tangent space is. But, if you work through the derivation of why the derivative is as such, you’ll see that you’re actually exploiting the fact that the determinant functional is a homomorphism.
