For your first question, let $X\subseteq M$ denote the fixed point set of $f$, and assume for a contradiction that $X$ contains infinitely many components. Label the first countably many of them $F_1, F_2,..$.
Now, create a sequence $x_i$ with $x_i\in F_i$. Because $M$ is compact, some subsequence of the $x_i$ converges to some $x\in M$. By abuse of notation, We'll assume the subsequence is the original subsequence and we'll write $\lim x_i = x$.
We first claim that $x\in X$. This is a simple consequence of continuity if $f$: $$f(x) = f(\lim x_i) = \lim f(x_i) = \lim x_i = x.$$
Now, let $U\subseteq M$ be a normal neighborhood of $x$, so that there is a unique minimal geodesic connecting $x$ to any point in $U$. Because $\lim x_i = x$, there are infinitely many $x_i\in U$. Suppose $x_j\in U$ is one of these.
Connecting $x$ to $x_j$ by a minimal geodesic $\gamma$, we find that $f\circ \gamma$ is a minimal geodesic connecting $x$ to $x_j$. By uniqueness, $f\circ \gamma = \gamma$, so $\gamma(t)$ is in $X$ for all $t$. Having a path from $x$ to $x_j$ in $X$, we deduce $x\in F_j$.
However, we can repeat this argument for a different $x_k\in U$, we find that $x\in F_j \cap F_k$ for $j\neq k$. This contradiction implies that there must have only been finitely many components.
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For the second question, suppose $f:M\rightarrow M$ is an isometry which preserves orientation. Let $F\subseteq M$ be a connected component of the fixed point set of $f$. I'm assuming you already know that $F$ is an embedded submanifold. We'll prove that $F$ has even codimension in $M$. That is, $\dim M - \dim F$ is an even number. (For your application, $M$ itself is even dimensional, which them implies $F$ is).
Let $p\in F$. The tangent space $T_p M$ has an orthgonal splitting $T_p M = T_p F \oplus V$ for some complementary subspace $V$. The goal is to show that $V$ has even dimension.
To that end, consider the differential $d_p f$. Because $d_p f$ obviously acts as the identity on $T_p F$, it follows that $d_p f$ preserves $V$ in the sense that $d_pf$ maps $V$ to itself by some isometry. Also, because $d_p f$ preserves the orientation and acts as the identity on $T_p F$, it must act via an orientation preserving isometry of $V$.
In addition, $d_p f$ cannot fix any non-zero $v\in V$, for otherwise the points $\exp(tv)$ are fixed by $f$, so $F$ would not have been a connected component of the fixed point set.
Because $d_p f:V\rightarrow V$ is an isometry, all its (complex) eigenvalues have modulus one. Those with non-zero imaginary part come in complex conjugate pairs. And by the previous paragraph, $1$ cannot be an eigenvalue of $d_p f:V\rightarrow V$. It follows that that number of $-1$-eigenvalues has the same parity as $\dim V$. In particular, $\det(d_p f:V\rightarrow V) = (-1)^{\dim V}$. Because $f$ preserves orientation, this determinant is positive, so $\dim V$ must be even.
