About Sylvester's determinant 
*

*If $A$ is any $n \times m$ matrix and $B$ is any $m \times n$ matrix then one familiar form of the Sylvester's identity is $\det(I + AB) = \det(I + BA)$. 
Now somehow curiously this above identity is often enough used along side another statement which says that for any vector $v$ and any invertible square matrix $A$, it is true that $\det(A + v v^T) = \det(A)\det(I + v^TA^{-1}v)$.
Is there some relationship between these two identities? Like can one be gotten from the other or vice versa? 

*A particularly useful (at least in some recent big breakthrough researches!) form of the Sylvester's identity is when $A= tu, B = v^\dagger $ where ($u$ and $v$ are complex vectors and $t$ is some complex number). This then shows that $\det(I + t u v^\dagger) = 1 + tv^\dagger u$
Now this simple statement above apparently implies this more powerful formulation which is not clear to me - If $C$ is any $n\times n$ (necessarily invertible?) matrix and $A$ is a rank-$1$ matrix then is $\det( I + tCA)$  is a degree $1$ polynomial in $t$? Why? (and this is the same as saying that this is "affine-linear" in t?) 
 A: The common ground of those two formulas is related to the Woodbury matrix identity. This relation is a useful statement that shows what happens to the inverse when one "updates" a matrix $A\in\mathbb{C}^{n\times n}$ with a term $UCV$, with $U,V\in\mathbb{C}^{n\times m}$, $C\in\mathbb{C}^{m\times m}$. The typical situation is when $m<n$ and this update is a small-rank term. The rank-1 version, often useful in its own, is called Sherman-Morrison formula. 
A statement on determinants can be derived with a small modification to the proof: take Equations (1) and (2) here; compute determinants of the first and third member of those chains of equalities, and equate them:
$$
\det A^{-1}\det(C-VA^{-1}U)^{-1} = \det\begin{bmatrix}A & U\\V&C\end{bmatrix}^{-1} = \det(A-UC^{-1}V)^{-1}\det C^{-1};
$$
invert everything and you get
$$
\det A \det(C-VA^{-1}U) = \det(A-UC^{-1}V)\det C, \tag{*}
$$
which by continuity holds also when $\begin{bmatrix}A & U\\V&C\end{bmatrix}$ isn't invertible. Your two formulas can be derived easily from this one ($C=I$ in both cases).
Equation (*) is also known as (the two versions of) the Schur complement formula; an equivalent statement appears for instance here (last equation of the section titled Background).
People in applied linear algebra often prefer talking about inverses than determinants, but it's essentially the same result.
A: For the first point, note that by Sylvester's identity
$$\det(I_n + v^TA^{-1}v) = \det(I_n + (v^T)(A^{-1}v)) = \det(I_n+(A^{-1}v)(v^T)) = \det(I_n + A^{-1}vv^T),$$
so 
$$\det(A)\det(I_n+v^TA^{-1}v) = \det(A)\det(I_n + A^{-1}vv^T) = \det(A + vv^T).$$
Given $u \in \mathbb{R}^m$ and $v \in \mathbb{R}^n$, their outer product is the $m\times n$ matrix $uv^T$. If $u$ and $v$ are non-zero, then their outer product has rank one. Conversely, a rank one $m\times n$ matrix can be written as the outer product of some non-zero $u$ and $v$.
Now suppose $C$ is an $n\times n$ matrix (not necessarily invertible) and $A$ is a rank one $n\times n$ matrix. By the above discussion, there are $u, v \in \mathbb{R}^n$ such that $A = uv^T$. So
\begin{align*}
\det(I_n + tCA) &= \det(I_n + tCuv^T)\\ 
&= \det(I_n + (tCu)(v^T))\\ 
&= \det(I_1 + (v^T)(tCu))\\ 
&= \det(I_1 + tv^TCu)\\ 
&= 1 + tv^TCu
\end{align*}
where the last equality follows because $I_1 + tv^TCu$ is a $1\times 1$ matrix.
A function of the form $f(t) = at + b$ is called affine. I'm guessing this is what is meant by affine-linear.
Added Later: Let me add a theorem which further demonstrates the relationship between the rank of a matrix and outer products. In the following statement, I am taking the definition of the rank of a matrix to be the dimension of its column space.

Theorem: An $m\times n$ matrix $A$ has rank $k$ if and only if  the minimal number of outer products needed to express $A$ as a sum is $k$ (i.e. $A$ can be written as the sum of $k$ outer products, but not $k - 1$).

Proving this is a really nice exercise in elementary linear algebra. I would hate to rob you of the experience by posting the solution here.
